High throughput Fizz Buzz

Fizz Buzz is a common challenge given during interviews. The challenge goes something like this:

Write a program that prints the numbers from 1 to n. If a number is divisible by 3, write Fizz instead. If a number is divisible by 5, write Buzz instead. However, if the number is divisible by both 3 and 5, write FizzBuzz instead.

The goal of this question is to write a FizzBuzz implementation that goes from 1 to infinity (or some arbitrary very very large number), and that implementation should do it as fast as possible.

Checking throughput

Write your fizz buzz program. Run it. Pipe the output through <your_program> | pv > /dev/null. The higher the throughput, the better you did.

Example

A naive implementation written in C gets you about 170MiB/s on an average machine:

#include <stdio.h>

int main() {
for (int i = 1; i < 1000000000; i++) {
if ((i % 3 == 0) && (i % 5 == 0)) {
printf("FizzBuzz\n");
} else if (i % 3 == 0) {
printf("Fizz\n");
} else if (i % 5 == 0) {
printf("Buzz\n");
} else {
printf("%d\n", i);
}
}
}


There is a lot of room for improvement here. In fact, I've seen an implementation that can get more than 3GiB/s on the same machine.

I want to see what clever ideas the community can come up with to push the throughput to its limits.

Rules

• All languages are allowed.
• The program output must be exactly valid fizzbuzz. No playing tricks such as writing null bytes in between the valid output - null bytes that don't show up in the console but do count towards pv throughput.

Here's an example of valid output:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz

# ... and so on


Valid output must be simple ASCII, single-byte per character, new lines are a single \n and not \r\n. The numbers and strings must be correct as per the FizzBuzz requirements. The output must go on forever (or a very high astronomical number) and not halt / change prematurely.

• Parallel implementations are allowed (and encouraged).

• Architecture specific optimizations / assembly is also allowed. This is not a real contest - I just want to see how people push fizz buzz to its limit - even if it only works in special circumstances/platforms.

Scores

Scores are from running on my desktop with an AMD 5950x CPU (16C / 32T). I have 32GB of 3600Mhz RAM.

**By far the best score so far is by @ais523 - generating FizzBuzz at a throughput that seems to average somewhere around 54-56GiB/s.

At the second place - @Paolo Bonzini with 40-41GiB/s.

Results for java:

Author Throughput
ioan2 2.6 GiB/s
randy 0.6 GiB/s
randy_ioan 2.4 GiB/s
ioan 4.5 GiB/s
olivier 6.0 GiB/s

Results for python:

Author Throughput
bconstanzo 0.1 GiB/s
michal 0.1 GiB/s
patrick 0.0 GiB/s
ksousa_chunking 0.1 GiB/s
ksousa_initial 0.0 GiB/s

Results for pypy:

Author Throughput
bconstanzo_pypy 0.3 GiB/s

Results for rust:

Author Throughput
aiden 3.4 GiB/s
xavier 0.9 GiB/s

Results for ruby:

Author Throughput
lonelyelk 0.1 GiB/s

Results for asm:

Author Throughput
ais523 56.6 GiB/s
paolo 41.6 GiB/s

Results for julia:

Author Throughput
marc 0.7 GiB/s

Results for c:

Author Throughput
random832 0.5 GiB/s
neil 4.8 GiB/s
kamila 4.1 GiB/s
xiver77 22.5 GiB/s
isaacg 1.9 GiB/s

Results for cpp:

Author Throughput
jdt 4.7 GiB/s
tobycpp 6.5 GiB/s

Plots generated using https://github.com/omertuc/fizzgolf

• How are we scoring speed? If you change the machine the speed will change drastically. You mentioned architecture specific optimizations as well, so it seems like it is tested on multiple different architectures. How do you ensure that comparison is fair? Nov 14 '20 at 18:07
• That's definitely a problem. I will try and run the implementations posted here on my desktop and keep score in the main question by editing it when new answers are posted. If it's something architecture specific I can't run, it'll still interesting to see, doesn't have to compete in the same score table. Nov 14 '20 at 18:09
• @agtoever note that the benchmark is done using piping to pv, which means that the Linux default pipe buffer size of 64K is used, as well as pv's default 64K buffer size. My entry exploits these properties by setting the buffer size to 64K, but it still blocks on write due to many factors; one of which is the generation speed being affected by the varying length of the printed numbers. Nov 15 '20 at 10:29
• related: My answer on an SO question about x86-64 asm fizz buzz includes a tutorial on tuning some parts of it for performance. (e.g. unroll by 3 and use a down-counter starting at 5 instead of modulo on i). But it still uses a slow div for int->string, and critically uses write for each group of 15 output lines built up in a buffer. Full-on perf tuning would probably just unroll by 15, and of course use much bigger buffers. Nov 16 '20 at 18:25
• Hoe do you measure throughput of a program which runs "for ever"? If I have a program which first does calculations for a year, then outputs a googol lines in a minute, it will lose if you stop measuring if you cut off the measurement within a year, but it'll be a winner if you measure for a year and a minute. This is of course an extreme example, but many programs will get slower when numbers get larger, so how you long you measure matters. Nov 21 '20 at 11:30

x86-64+AVX2 assembly language (Linux, cpp+gas)

Build and usage instructions

This program is most conveniently built using gcc. Save it as fizzbuzz.S (that's a capital S as the extension), and build using the commands

gcc -mavx2 -c fizzbuzz.S
ld -o fizzbuzz fizzbuzz.o


Run as ./fizzbuzz piped into one command, e.g. ./fizzbuzz | pv > /dev/null (as suggested in the question), ./fizzbuzz | cat, or ./fizzbuzz | less. To simplify the I/O, this will not work (producing an error on startup) if you try to output to a file/terminal/device rather than a pipe. Additionally, this program may produce incorrect output if piped into two commands (e.g. ./fizzbuzz | pv | cat > fizzbuzz.txt), but only in the case where the middle command uses the splice system call; this is either a bug in Linux (very possible with system calls this obscure!) or a mistake in the documentation of the system calls in question (also possible). However, it should work correctly for the use case in the question, which is all that matters on CGCC.

This program is somewhat system-specific; it requires the operating system to be a non-ancient version of Linux, and the processor to be an x86-64 implementation that supports AVX2. (Most moderately recent processors by Intel and AMD have AVX2 support, including the Ryzen 9 mentioned in the question, and almost all use the x86-64 instruction set.) However, it avoids assumptions about the system it's running on beyond those mentioned in the header, so there's a decent chance that if you can run Linux, you can run this.

The program outputs a quintillion lines of FizzBuzz and then exits (going further runs into problems related to the sizes of registers). This would take tens of years to accomplish, so hopefully counts as "a very high astronomical number" (although it astonishes me that it's a small enough timespan that it might be theoretically possible to reach a number as large as a quintillion without the computer breaking).

As a note: this program's performance is dependent on whether it and the program it outputs to are running on sibling CPUs or not, something which will be determined arbitrarily by the kernel when you start it. If you want to compare the two possible timings, use taskset to force the programs onto particular CPUs: taskset 1 ./fizzbuzz | taskset 2 pv > /dev/null versus taskset 1 ./fizzbuzz | taskset 4 pv > /dev/null. (The former will probably run faster, but might be slower on some CPU configurations.)

Discussion

I've spent months working on this program. I've long thought that "how fast can you make a FizzBuzz" would be a really interesting question for learning about high-performance programming, and when I subsequently saw this question posted on CGCC, I pretty much had to try.

This program aims for the maximum possible single-threaded performance. In terms of the FizzBuzz calculation itself, it is intended to sustain a performance of 64 bytes of FizzBuzz per 4 clock cycles (and is future-proofed where possible to be able to run faster if the relevant processor bottleneck – L2 cache write speed – is ever removed). This is faster than a number of standard functions. In particular, it's faster than memcpy, which presents interesting challenges when it comes to I/O (if you try to output using write then the copies in write will take up almost all the runtime – replacing the I/O routine here with write causes the performance on my CPU to drop by a factor of 5). As such, I needed to use much more obscure system calls to keep I/O-related copies to a minimum (in particular, the generated FizzBuzz text is only sent to main memory if absolutely necessary; most of the time it's stored in the processor's L2 cache and piped into the target program from there, which is why reading it from a sibling CPU can boost performance – the physical connection to the L2 cache is shorter and higher bandwidth than it would be to a more distant CPU).

On my computer (which has a fairly recent, but not particularly powerful, Intel processor), this program generates around 31GiB of FizzBuzz per second. I'll be interested to see how it does on the OP's computer.

I did experiment with multithreaded versions of the program, but was unable to gain any speed. Experiments with simpler programs show that it could be possible, but any gains may be small; the cost of communication between CPUs is sufficiently high to negate most of the gains you could get by doing work in parallel, assuming that you only have one program reading the resulting FizzBuzz (and anything that writes to memory will be limited by the write speed of main memory, which is slower than the speed with which the FizzBuzz can be generated).

The program

This isn't , so my explanation of the program and its algorithm are given as comments in the program itself. (I still had to lightly golf the program, and especially the explanation, to fit this post within the 65536 byte size limit.)

The program is written in a "literate" assembly style; it will be easiest to understand if you read it in order, from start to end. (I also added a number of otherwise useless line labels to separate the program into logical groups of instructions, in order to make the disassembly easier to read, if you're one of the people who prefers to read assembly code like that.)

.intel_syntax prefix

#include <asm/errno.h>
#include <asm/mman.h>
#include <asm/unistd.h>
#define F_SETPIPE_SZ 1031 // not in asm headers, define it manually

// The Linux system call API (limited to 4 arguments, the most this
// program uses). 64-bit registers are unsuffixed; 32-bit have an "e"
// suffix.
#define ARG1 %rdi
#define ARG1e %edi
#define ARG2 %rsi
#define ARG2e %esi
#define ARG3 %rdx
#define ARG3e %edx
#define ARG4 %r10
#define ARG4e %r10d
#define SYSCALL_RETURN %rax
#define SYSCALL_RETURNe %eax
#define SYSCALL_NUMBER %eax

// %rax, %rcx, %rdx, %ymm0-3 are general-purpose temporaries. Every
// other register is used for just one or two defined purposes; define
// symbolic names for them for readability. (Bear in mind that some of
// these will be clobbered sometimes, e.g. OUTPUT_LIMIT is clobbered
// by syscall because it's %r11.)
#define OUTPUT_PTR %rbx
#define BYTECODE_IP %rbp
#define SPILL %rsi
#define BYTECODE_GEN_PTR %rdi
#define REGEN_TRIGGER %r8
#define REGEN_TRIGGERe %r8d
#define YMMS_AT_WIDTH %r9
#define YMMS_AT_WIDTHe %r9d
#define BUZZ %r10
#define BYTECODE_NEG_LEN %r10
#define FIZZ %r11
#define FIZZe %r11d
#define OUTPUT_LIMIT %r11
#define BYTECODE_END %r12
#define BYTECODE_START %r13
#define BYTECODE_STARTe %r13d
#define PIPE_SIZE %r13
#define LINENO_WIDTH %r14
#define LINENO_WIDTHe %r14d
#define GROUPS_OF_15 %r15
#define GROUPS_OF_15e %r15d
#define LINENO_LOW %ymm4
#define LINENO_MID %ymm5
#define LINENO_MIDx %xmm5
#define LINENO_TOP %ymm6
#define LINENO_TOPx %xmm6
#define LINENO_MID_TEMP %ymm7
#define ENDIAN_SHUFFLE %ymm8
#define ENDIAN_SHUFFLEx %xmm8
#define LINENO_LOW_INCR %ymm9
#define LINENO_LOW_INCRx %xmm9

// The last six vector registers are used to store constants, to avoid
#define LINENO_LOW_INIT %ymm10
#define LINENO_MID_BASE %ymm11
#define LINENO_TOP_MAX %ymm12
#define ASCII_OFFSET %ymm13
#define ASCII_OFFSETx %xmm13
#define BIASCII_OFFSET %ymm14
#define BASCII_OFFSET %ymm15

// Global variables.
.bss
.align 4 << 20
// The most important global variables are the IO buffers. There are
// two of these, each with 2MiB of memory allocated (not all of it is
// used, but putting them 2MiB apart allows us to simplify the page
// table; this gives a 30% speedup because page table contention is
// one of the main limiting factors on the performance).
io_buffers:
.zero 2 * (2 << 20)
// The remaining 2MiB of memory stores everything else:
iovec_base:          // I/O config buffer for vmsplice(2) system call
.zero 16
error_write_buffer:  // I/O data buffer for write(2) system call
.zero 1
.p2align 9,0
bytecode_storage:    // the rest is a buffer for storing bytecode
.zero (2 << 20) - 512

// The program starts here. It doesn't use the standard library (or
// indeed any libraries), so the start point is _start, not main.
.text
.globl _start
_start:

// This is an AVX2 program, so check for AVX2 support by running an
// AVX2 command. This is a no-op, but generates SIGILL if AVX2 isn't
// supported.
vpand %ymm0, %ymm0, %ymm0

// Initialize constant registers to their constant values.
vmovdqa LINENO_LOW_INIT, [%rip + lineno_low_init]
vmovdqa LINENO_MID_BASE, [%rip + lineno_mid_base]
vmovdqa LINENO_TOP_MAX, [%rip + lineno_top_max]
vmovdqa ASCII_OFFSET, [%rip + ascii_offset]
vmovdqa BIASCII_OFFSET, [%rip + biascii_offset]
vmovdqa BASCII_OFFSET, [%rip + bascii_offset]

// Initialize global variables to their initial values.
vmovdqa ENDIAN_SHUFFLE, [%rip + endian_shuffle_init]
vmovdqa LINENO_TOP, [%rip + lineno_top_init]

// Check the size of the L2 cache.
//
// This uses the CPUID interface. To use it safely, check what range
// of command numbers is legal; commands above the legal range have
// undefined behaviour, commands within the range might not be
// implemented but will return all-zeros rather than undefined values.
// CPUID clobbers a lot of registers, including some that are normally
// call-preserved, so this must be done first.
mov %eax, 0x80000000 // asks which CPUID extended commands exist
cpuid                // returns the highest supported command in %eax
cmp %eax, 0x80000006 // does 0x80000006 give defined results?

cpuid                // returns size in KiB in the top half of %ecx
shr %ecx, 16
jz bad_cpuid_error   // unsupported commands return all-0s

// Calculate the desired pipe size, half the size of the L2 cache.
// This value is chosen so that the processor can hold a pipeful of
// data being output, plus a pipeful of data being calculated, without
// needing to resort to slow L3 memory operations.
shl %ecx, 10 - 1     // convert KiB to bytes, then halve
mov PIPE_SIZE, %rcx

// Ask the kernel to resize the pipe on standard output.
mov ARG1e, 1
mov ARG2e, F_SETPIPE_SZ
mov ARG3e, %ecx
mov SYSCALL_NUMBER, __NR_fcntl
syscall
je pipe_error
cmp SYSCALL_RETURNe, -EPERM
je pipe_perm_error
call exit_on_error
cmp SYSCALL_RETURN, PIPE_SIZE
jne pipe_size_mismatch_error

// Ask the kernel to defragment the physical memory backing the BSS
// (read-write data) segment. This simplifies the calculations needed
// to find physical memory addresses, something that both the kernel
// and processor would otherwise spend a lot of time doing, and
// speeding the program up by 30%.
lea ARG1, [%rip + io_buffers]
mov ARG2e, 3 * (2 << 20)
syscall
call exit_on_error

// From now on, OUTPUT_PTR is permanently set to the memory location
// where the output is being written. This starts at the start of the
// first I/O buffer.
lea OUTPUT_PTR, [%rip + io_buffers]

///// First phase of output
//
// The FizzBuzz output is produced in three distinct phases. The first
// phase is trivial; just a hardcoded string, that's left in the
// output buffer, to be output at the end of the second phase.

first_phase:

.section .rodata
fizzbuzz_intro:
.ascii "1\n2\nFizz\n4\nBuzz\nFizz\n7\n8\nFizz\n"
.text
vmovdqu %ymm0, [%rip + fizzbuzz_intro]
vmovdqu [OUTPUT_PTR], %ymm0

///// Second phase of output
//
// This is a routine implementing FizzBuzz in x86-64+AVX2 assembler in
// a fairly straightforward and efficient way. This isn't as fast as
// the third-phase algorithm, and can't handle large numbers, but will
// introduce some of the basic techniques this program uses.

second_phase_init:

// The outer loop of the whole program breaks the FizzBuzz output into
// sections where all the line numbers contain the same number of
// digits. From now on, LINENO_WIDTH tracks the number of digits in
// the line number. This is currently 2; it ranges from 2-digit
// numbers to 18-digit numbers, and then the program ends.
mov LINENO_WIDTHe, 2

// GROUPS_OF_15 is permanently set to the number of groups of 15 lines
// that exist at this line number width; it's multiplied by 10 whenever
// LINENO_WIDTH is incremented.
//
// A general note about style: often the program uses numbers that are
// statically known to fit into 32 bits, even in a register that's
// conceptually 64 bits wide (like this one). In such cases, the
// 32-bit and 64-bit versions of a command will be equivalent (as the
// 32-bit version zero-extends to 64-bits on a 64-bit processor); this
// program generally uses the 32-bit version, both because it
// sometimes encodes to fewer bytes (saving cache pressure), and
// because some processors recognise zeroing idioms only if they're 32
// bits wide.
mov GROUPS_OF_15e, 6

// Some constants used throughout the second phase, which permanently
// stay in their registers. Note that short string literals can be
// stored in normal integer registers - the processor doesn't care.
mov FIZZ, 0x0a7a7a6946  // "Fizz\n"
mov BUZZ, 0x0a7a7a7542  // "Buzz\n"

.section .rodata
.p2align 5, 0
second_phase_constants:
.byte 0, 0, 0, 0, 0, 0, 0, 0
.byte 1, 0, 0, 0, 0, 0, 0, 0
.text
vmovdqa %xmm3, [%rip + second_phase_constants]

// This program makes extensive use of a number format that I call
// "high-decimal". This is a version of decimal where the digit 0 is
// encoded as the byte 246, the digit 1 as the byte 247, ..., the
// digit 9 as the byte 255. The bytes are stored in the normal
// endianness for the processor (i.e. least significant first), and
// padded to a known length (typically 8 digits) with leading zeroes.
//
// The point of high-decimal is that it allows us to use arithmetic
// operators intended for binary on high-decimal numbers, and the
// carries will work the same way (i.e. the same digits will carry,
// although carries will be 0-based rather than 246-based); all that's
// required is to identify the digits that carried and add 246 to
// them. That means that the processor's binary ALU can be used to do
// additions directly in decimal - there's no need for loops or
// anything like that, and no need to do binary/decimal conversions.
//
// The first use for high-decimal is to store the line number during
// the second phase (it's stored differently in the third phase).
// It's stored it in the top half of %xmm1 (although it's only 64 bits
// wide, it needs to be in a vector register so that it can be
// interpreted as 8 x 8 bits when necessary; general-purpose
// registers can't do that). The bottom half of %xmm1 is unused, and
// frequently overwritten with arbitrary data.
.section .rodata
line_number_init:
#define REP8(x) x,x,x,x,x,x,x,x
.byte REP8(0)
.byte 246, 247, 246, 246, 246, 246, 246, 246
.text
vmovdqa %xmm1, [%rip + line_number_init]

// Writing line numbers is nontrivial because x86-64 is little-endian
// but FizzBuzz output is big-endian; also, leading zeroes aren't
// allowed. ENDIAN_SHUFFLE is used to fix both these problems; when
// used to control the vector shuffler, it reverses the order of a
// vector register, and rotates the elements to put the first digit
// (based on LINENO_WIDTH) into the first byte. (This method is used
// by both the second and third phases; the second phase uses only the
// bottom half, with the top half used by the third phase, but they
// are both initialized together.)
.section .rodata
endian_shuffle_init:
.byte 9, 8, 7, 6, 5, 4, 3, 2
.byte 1, 0, 255, 254, 253, 252, 251, 250
.byte 3, 2, 1, 0, 255, 254, 253, 252
.byte 251, 250, 249, 248, 247, 246, 245, 244
.text

second_phase_per_width_init:

// The second phase writing routines are macros.
//
// Fizz and Buzz are trivial. (This writes a little beyond the end of
// the string, but that's OK; the next line will overwrite them.)
#define WRITE_FIZZ   mov [OUTPUT_PTR], FIZZ; add OUTPUT_PTR, 5
#define WRITE_BUZZ   mov [OUTPUT_PTR], BUZZ; add OUTPUT_PTR, 5

// For FizzBuzz, output 32 bits of FIZZ to write "Fizz" with no
// newline, then write a "Buzz" after that.
#define WRITE_FIZZBUZZ \
mov [OUTPUT_PTR], FIZZe; mov [OUTPUT_PTR + 4], BUZZ; \

// To write a line number, add 58 to each byte of the line number
// %xmm1, fix the endianness and width with a shuffle, and write a
// final newline.
.section .rodata
ascii_offset:
.byte REP8(58), REP8(58), REP8(58), REP8(58)
.text
#define WRITE_LINENO \
vpshufb %xmm0, %xmm0, ENDIAN_SHUFFLEx; \
vmovdqu [OUTPUT_PTR], %xmm0; \
lea OUTPUT_PTR, [OUTPUT_PTR + LINENO_WIDTH + 1]; \
mov byte ptr [OUTPUT_PTR - 1], 10  // 10 = newline

// Incrementing the line number is fairly easy: add 1 (in the usual
// binary notation, taken from %xmm3) to the high-decimal number, then
// convert any bytes that produced a carry to high-decimal 0s by
// max-ing with 246.
//
// Normally I'd use a separate constant for this, but there randomly
// happens to be an %xmm register with 246s in its top half already
// (it's intended for an entirely different purpose, but it'll do for
// this one too).
#define INC_LINENO \
vpaddq %xmm1, %xmm3, %xmm1; vpmaxub %xmm1, LINENO_TOPx, %xmm1

// Avoid modulus tests by unrolling the FizzBuzz by 15. (Bear in mind
// that this starts at 10, not 0, so the pattern will have a different
// phase than usual.)
mov %ecx, GROUPS_OF_15e
fifteen_second_phase_fizzbuzz_lines:
WRITE_BUZZ; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_FIZZ; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_FIZZBUZZ; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_FIZZ; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_BUZZ; INC_LINENO
WRITE_FIZZ; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_LINENO; INC_LINENO
WRITE_FIZZ; INC_LINENO
dec %ecx
jnz fifteen_second_phase_fizzbuzz_lines

second_phase_increment_width:

lea GROUPS_OF_15e, [GROUPS_OF_15 + GROUPS_OF_15 * 4]
inc LINENO_WIDTHe

// Increment every element of the low half of ENDIAN_SHUFFLE to
// adjust it for the new width, while leaving the top half unchanged.
vpcmpeqb %xmm0, %xmm0, %xmm0
vpsubb ENDIAN_SHUFFLE, ENDIAN_SHUFFLE, %ymm0

// The second phase handles line numbers with 2 to 5 digits.
cmp LINENO_WIDTHe, 6
jne second_phase_per_width_init

///// The output routine
//
// Most FizzBuzz routines produce output with write or a similar
// system call, but these have the disadvantage that they need to copy
// the data being output from userspace into kernelspace. It turns out
// that when running full speed (as seen in the third phase), FizzBuzz
// actually runs faster than memcpy does, so write and friends are
// unusable when aiming for performance - this program runs five times
// faster than an equivalent that uses write-like system calls.
//
// To produce output without losing speed, the program therefore needs
// to avoid copies, or at least do them in parallel with calculating
// the next block of output. This can be accomplished with the
// vmsplice system call, which tells the kernel to place a reference
// to a buffer into a pipe (as opposed to copying the data into the
// pipe); the program at the other end of this pipe will then be able
// to read the output directly out of this program's memory, with no
// need to copy the data into kernelspace and then back into
// userspace. In fact, it will be reading out of this program's
// processor's L2 cache, without main memory being touched at all;
// this is the secret to high-performance programming, because the
// cache is much faster than main memory is.
//
// Of course, it's therefore important to avoid changing the output
// buffer until the program connected to standard output has actually
// read it all. This is why the pipe size needed to be set earlier; as
// long as the amount of output is always at least as large as the
// pipe size, successfully outputting one buffer will ensure that none
// of the other buffer is left in the pipe, and thus it's safe to
// overwrite the memory that was previously output. There is some need
// to jump through hoops later on to make sure that swap_buffers is
// never called with less than one pipeful of data, but it's worth it
// to get the huge performance boost.

mov %rdx, OUTPUT_PTR
and %edx, (2 << 20) - 1

call swap_buffers
jmp third_phase_init

// Takes the amount of data to output in %rdx, and outputs from the
// buffer containing OUTPUT_PTR.
swap_buffers:
and OUTPUT_PTR, -(2 << 20)  // rewind to the start of the buffer
mov [%rip + iovec_base], OUTPUT_PTR
mov [%rip + iovec_base + 8], %rdx
mov ARG1e, 1
lea ARG2, [%rip + iovec_base]
mov ARG3e, 1
xor ARG4e, ARG4e

// As with most output commands, vmsplice can do a short write
// sometimes, so it needs to be called in a loop in order to ensure
// that all the output is actually sent.
1: mov SYSCALL_NUMBER, __NR_vmsplice
syscall
call exit_on_error
sub [ARG2 + 8], SYSCALL_RETURN
jnz 1b

xor OUTPUT_PTR, (2 << 20)  // swap to the other buffer
ret

///// Third phase of output
//
// This is the heart of this program. It aims to be able to produce a
// sustained output rate of 64 bytes of FizzBuzz per four clock cycles
// in its main loop (with frequent breaks to do I/O, and rare breaks
// to do more expensive calculations).
//
// The third phase operates primarily using a bytecode interpreter; it
// generates a program in "FizzBuzz bytecode", for which each byte of
// bytecode generates one byte of output. The bytecode language is
// designed so that it can be interpreted using SIMD instructions; 32
// bytes of bytecode can be loaded from memory, interpreted, and have
// its output stored back into memory using just four machine
// instructions. This makes it possible to speed up the FizzBuzz
// calculations by hardcoding some of the calculations into the
// bytecode (this is similar to how JIT compilers can create a version
// of the program with some variables hardcoded, and throw it away on
// the rare occasions that those variables' values change).

third_phase_init:

// Reinitialize ENDIAN_SHUFFLE by copying the initializer stored in
// its high half to both halves. This works in the same way as in the
// second phase.
vpermq ENDIAN_SHUFFLE, ENDIAN_SHUFFLE, 0xEE

// Up to this point, PIPE_SIZE has held the size of the pipe. In order
// to save on registers, the pipe size is from now on encoded via the
// location in which the bytecode program is stored; the bytecode is
// started at iovec_base + PIPE_SIZE (which will be somewhere within
// bytecode_storage), so the same register can be used to find the
// bytecode and to remember the pipe size.
lea %rax, [%rip + iovec_base]
add BYTECODE_START, %rax  // BYTECODE_START is a synonym for PIPE_SIZE

// The bytecode program always holds instructions to produce exactly
// 600 lines of FizzBuzz. At width 6, those come to 3800 bytes long.
lea BYTECODE_END, [BYTECODE_START + 3800]

mov REGEN_TRIGGER, -1  // irrelevant until much later, explained there

third_phase_per_width_init:

// Calculate the amount of output at this LINENO_WIDTH. The result
// will always be divisible by 32, and thus is stored as the number of
// 32-byte units at this width; storing it in bytes would be more
// convenient, but sadly would overflow a 64-bit integer towards the
// end of the program.
lea %ecx, [LINENO_WIDTH * 8 + 47]   // bytes per 15 lines
mov YMMS_AT_WIDTH, GROUPS_OF_15
shr YMMS_AT_WIDTH, 5   // to avoid overflow, divide by 32 first
imul YMMS_AT_WIDTH, %rcx

// This program aims to output 64 bytes of output per four clock
// cycles, which it achieves via a continuous stream of 32-byte writes
// calculated by the bytecode program. One major complication here is
// that the 32-byte writes won't correspond to lines of FizzBuzz; a
// single processor instruction may end up outputting multiple
// different line numbers. So it's no longer possible to have a simple
// line number register, like it was in the second phase.
//
// Instead, the program stores an *approximation* of the line number,
// which is never allowed to differ by 100 or more from the "actual"
// line number; the bytecode program is responsible for fixing up the
// approximation to work out the correct line number to output (this
// allows the same CPU instruction to output digits from multiple
// different line numbers, because the bytecode is being interpreted
// in a SIMD way and thus different parts of the bytecode can fix the
// line number up differently within a single instruction.
//
// The line number is split over three processor registers:
// - LINENO_LOW: stores the line number modulo 200
// - LINENO_MID: stores the hundreds to billions digits
// - LINENO_TOP: stores the ten-billions and more significant digits
// (The parity of the 100s digit is duplicated between LINENO_MID and
// LINENO_LOW; this allows a faster algorithm for LINENO_MID updates.)
//
// Because there's only a need to be within 100 of the real line
// number, the algorithm for updating the line numbers doesn't need to
// run all that often (saving processor cycles); it runs once every
// 512 bytes of output, by simply adding a precalculated value
// (LINENO_LOW_INCR) to LINENO_LOW, then processing the carry to
// LINENO_MID (see later for LINENO_TOP). The amount by which the line
// number increases per 512 bytes of output is not normally going to
// be an integer; LINENO_LOW is therefore stored as a 64-bit fixpoint
// number (in which 2**64 represents "200", e.g. 2**63 would be the
// representation of "the line number is 100 mod 200"), in order to
// delay the accumulation of rounding errors as long as possible. It's
// being stored in a vector register, so there are four copies of its
// value; two of them have 50 (i.e 2**62) added, and two of them have
// 50 subtracted, in order to allow for more efficient code to handle
// the carry to LINENO_MID. Additionally, LINENO_LOW is interpreted as
// a signed number (an older version of this program was better at
// checking for signed than unsigned overflow and I had no reason to
// change).
//
// LINENO_LOW and LINENO_MID are reset every LINENO_WIDTH increase
// (this is because the program can calculate "past" the width
// increase due to not being able to break out of every instruction of
// the main loop, which may cause unwanted carries into LINENO_MID and
// force a reset).

.section .rodata
lineno_low_init:
.byte 0, 0, 0, 0, 0, 0, 0, 192
.byte 0, 0, 0, 0, 0, 0, 0, 64
.byte 0, 0, 0, 0, 0, 0, 0, 192
.byte 0, 0, 0, 0, 0, 0, 0, 64
.text
vmovdqa LINENO_LOW, LINENO_LOW_INIT

// %ecx is the number of bytes in 15 lines. That means that the number
// of 200-line units in 512 bytes is 38.4/%ecx, i.e. 384/(%ecx*10).
// Multiply by 2**64 (i.e. 384*2**64/(%ecx*10) to get LINENO_LOW_INCR.
lea %ecx, [%rcx + %rcx * 4]
mov %edx, 384
xor %eax, %eax
div %rcx  // 128-bit divide, %rax = %rdx%rax / %rcx
vpxor LINENO_LOW_INCR, LINENO_LOW_INCR, LINENO_LOW_INCR
vpinsrq LINENO_LOW_INCRx, LINENO_LOW_INCRx, %rax, 0
vpermq LINENO_LOW_INCR, LINENO_LOW_INCR, 0

// LINENO_MID is almost stored in high-decimal, as four eight-digit
// numbers. However, the number represented is the closest line number
// that's 50 mod 100, stored as the two closest multiples of 100 (e.g.
// if the true line number is 235, it's approximated as 250 and then
// stored using the representations for 200 and 300), which is why
// LINENO_LOW needs the offsets of 50 and -50 to easily do a carry. A
// ymm vector holds four 64-bit numbers, two of which hold the value
// that's 0 mod 200, two which hold the value that's 100 mod 200. So
// carries on it are handled using a vector of mostly 246s, with 247s
// in the two locations which are always odd.
.section .rodata
lineno_mid_base:
.byte 246, 246, 246, 246, 246, 246, 246, 246
.byte 247, 246, 246, 246, 246, 246, 246, 246
.byte 246, 246, 246, 246, 246, 246, 246, 246
.byte 247, 246, 246, 246, 246, 246, 246, 246
.text

// This code is some fairly complex vector manipulation to initialise
// LINENO_MID to a power of 10 (handling the case where LINENO_WIDTH
// is so high that the hundreds to billions digits are all zeroes).
mov %edx, 1
mov %eax, 11
sub %eax, LINENO_WIDTHe
cmovbe %eax, %edx
shl %eax, 3
vpxor %xmm0, %xmm0, %xmm0
vpinsrq %xmm0, %xmm0, %rax, 0
vpermq %ymm0, %ymm0, 0
vpcmpeqb LINENO_MID, LINENO_MID, LINENO_MID
vpsrlq LINENO_MID, LINENO_MID, %xmm0
vpmaxub LINENO_MID, LINENO_MID_BASE, LINENO_MID
vpermq %ymm0, LINENO_MID_BASE, 0x55
vpsubb %ymm0, %ymm0, LINENO_MID_BASE
vpmaxub LINENO_MID, LINENO_MID_BASE, LINENO_MID

// LINENO_TOP doesn't need to be initialized for new widths, because
// an overrun by 100 lines is possible, but by 10 billion lines isn't.
// The format consists of two 64-bit sections that hold high-decimal
// numbers (these are always the same as each other), and two that
// hold constants that are used by the bytecode generator.
.section .rodata
lineno_top_init:
.byte 198, 197, 196, 195, 194, 193, 192, 191
.byte 246, 246, 246, 246, 246, 246, 246, 246
.byte 190, 189, 188, 187, 186, 185, 184, 183
.byte 246, 246, 246, 246, 246, 246, 246, 246
.text

// When moving onto a new width, start at the start of the bytecode
// program.
mov BYTECODE_IP, BYTECODE_START

// Generating the bytecode program
//
// The bytecode format is very simple (in order to allow it to be
// interpreted in just a couple of machine instructions):
// - A negative byte represents a literal character (e.g. to produce
//   a literal 'F', you use the bytecode -'F', i.e. -70 = 0xba)
// - A byte 0..7 represents the hundreds..billions digit of the line
//   number respectively, and asserts that the hundreds digit of the
//   line number is even
// - A byte 8..15 represents the hundreds..billions digit of the line
//   number respectively, and asserts that the hundreds digit of the
//   line number is odd
//
// In other words, the bytecode program only ever needs to read from
// LINENO_MID; the information stored in LINENO_LOW and LINENO_TOP
// therefore has to be hardcoded into it. The program therefore needs
// to be able to generate 600 lines of output (as the smallest number
// that's divisible by 100 to be able to hardcode the two low digits,
// 200 to be able to get the assertions about the hundreds digits
// correct, and 3 and 5 to get the Fizzes and Buzzes in the right
// place).

generate_bytecode:

mov BYTECODE_GEN_PTR, BYTECODE_START

// FIZZ and BUZZ work just like in the second phase, except that they
// are now bytecode programs rather than ASCII.
mov FIZZ, 0xf6868697ba  // -"Fizz\n"
mov BUZZ, 0xf686868bbe  // -"Buzz\n"

// %ymm2 holds the bytecode for outputting the hundreds and more
// significant digits of a line number. The most significant digits of
// this can be obtained by converting LINENO_TOP from high-decimal to
// the corresponding bytecode, which is accomplished by subtracting
// from 198 (i.e. 256 - 10 - '0'). The constant parts of LINENO_TOP
// are 198 minus the bytecode for outputting the hundreds to billions
// digit of a number; this makes it possible for a single endian
// shuffle to deal with all 16 of the mid and high digits at once.
.section .rodata
bascii_offset:
.byte REP8(198), REP8(198), REP8(198), REP8(198)
.text
vpsubb %ymm2, BASCII_OFFSET, LINENO_TOP
vpshufb %ymm2, %ymm2, ENDIAN_SHUFFLE

#define GEN_FIZZ  mov [BYTECODE_GEN_PTR], FIZZ; add BYTECODE_GEN_PTR, 5
#define GEN_BUZZ  mov [BYTECODE_GEN_PTR], BUZZ; add BYTECODE_GEN_PTR, 5
#define GEN_FIZZBUZZ \
mov [BYTECODE_GEN_PTR], FIZZe; \
mov [BYTECODE_GEN_PTR + 4], BUZZ; add BYTECODE_GEN_PTR, 9
#define GEN_LINENO(units_digit) \
vmovdqu [BYTECODE_GEN_PTR], %xmm2; \
lea BYTECODE_GEN_PTR, [BYTECODE_GEN_PTR + LINENO_WIDTH + 1]; \
mov [BYTECODE_GEN_PTR - 3], %al; \
mov word ptr [BYTECODE_GEN_PTR - 2], 0xf6d0 - units_digit

// The bytecode generation loop is unrolled to depth 30, allowing the
// units digits to be hardcoded. The tens digit is stored in %al, and
// incremented every ten lines of output. The parity of the hundreds
// digit is stored in %ymm2: one half predicts the hundreds digit to
// be even, the other to be odd, and the halves are swapped every time
// the tens digit carries (ensuring the predictions are correct).
mov %eax, 0xd0
jmp 2f
inc_tens_digit:
cmp %al, 0xc7
je 1f  // jumps every 10th execution, therefore predicts perfectly
dec %eax
ret
1: mov %eax, 0xd0
vpermq %ymm2, %ymm2, 0x4e
ret

2: mov %ecx, 20
thirty_bytecode_lines:
GEN_BUZZ
GEN_LINENO(1)
GEN_FIZZ
GEN_LINENO(3)
GEN_LINENO(4)
GEN_FIZZBUZZ
GEN_LINENO(6)
GEN_LINENO(7)
GEN_FIZZ
GEN_LINENO(9)
call inc_tens_digit
GEN_BUZZ
GEN_FIZZ
GEN_LINENO(2)
GEN_LINENO(3)
GEN_FIZZ
GEN_BUZZ
GEN_LINENO(6)
GEN_FIZZ
GEN_LINENO(8)
GEN_LINENO(9)
call inc_tens_digit
GEN_FIZZBUZZ
GEN_LINENO(1)
GEN_LINENO(2)
GEN_FIZZ
GEN_LINENO(4)
GEN_BUZZ
GEN_FIZZ
GEN_LINENO(7)
GEN_LINENO(8)
GEN_FIZZ
call inc_tens_digit
dec %ecx
jnz thirty_bytecode_lines

generate_bytecode_overrun_area:

// Duplicate the first 512 bytes of the bytecode program at the end,
// so that there's no need to check to see whether BYTECODE_IP needs
// to be looped back to the start of the program any more than once
// per 512 bytes
mov %rax, BYTECODE_START
#define COPY_64_BYTECODE_BYTES(offset) \
vmovdqa %ymm0, [%rax + offset]; \
vmovdqa %ymm3, [%rax + (offset + 32)]; \
vmovdqu [BYTECODE_GEN_PTR + offset], %ymm0; \
vmovdqu [BYTECODE_GEN_PTR + (offset + 32)], %ymm3
COPY_64_BYTECODE_BYTES(0)
COPY_64_BYTECODE_BYTES(64)
COPY_64_BYTECODE_BYTES(128)
COPY_64_BYTECODE_BYTES(192)
COPY_64_BYTECODE_BYTES(256)
COPY_64_BYTECODE_BYTES(320)
COPY_64_BYTECODE_BYTES(384)
COPY_64_BYTECODE_BYTES(448)

// Preparing for the main loop
//
// Work out how long the main loop is going to iterate for.
// OUTPUT_LIMIT holds the address just beyond the end of the output
// that the main loop should produce. The aim here is to produce
// exactly one pipeful of data if possible, but to stop earlier if
// there's a change in digit width (because any output beyond that
// point will be useless: the bytecode will give it the wrong number
// of digits).
calculate_main_loop_iterations:

// Extract the pipe size from BYTECODE_START, in 32-byte units.
// During this calculation, OUTPUT_LIMIT holds the amount of output
// produced, rather than an address like normal.
mov OUTPUT_LIMIT, BYTECODE_START
lea %rdx, [%rip + iovec_base]
sub OUTPUT_LIMIT, %rdx
shr OUTPUT_LIMIT, 5

// Reduce the output limit to the end of this width, if it would be
// higher than that.
cmp OUTPUT_LIMIT, YMMS_AT_WIDTH
cmovae OUTPUT_LIMIT, YMMS_AT_WIDTH

// If there's already some output in the buffer, reduce the amount
// of additional output produced accordingly (whilst ensuring that
// a multiple of 512 bytes of output is produced).
//
// This would be buggy if the YMMS_AT_WIDTH limit were hit at the
// same time, but that never occurs as it would require two width
// changes within one pipeful of each other, and 9000000 lines of
// FizzBuzz is much more than a pipeful in size.
mov %rax, OUTPUT_PTR
and %eax, ((2 << 20) - 1) & -512
shr %eax, 5
sub OUTPUT_LIMIT, %rax

// The amount of output to produce is available now, and won't be
// later, so subtract it from the amount of output that needs to
// be produced now.
sub YMMS_AT_WIDTH, OUTPUT_LIMIT

// Return OUTPUT_LIMIT back to being a pointer, not an amount.
shl OUTPUT_LIMIT, 5

prepare_main_loop_invariants:

// To save one instruction in the bytecode interpreter (which is very
// valuable, as it runs every second CPU cycle), LINENO_MID_TEMP is
// used to store a reformatted version of LINENO_MID, in which each
// byte is translated from high-decimal to ASCII, and the bytecode
// command that would access that byte is added to the result (e.g.
// the thousands digit for the hundreds-digits-odd version has 10
// added to convert from high-decimal to a pure number, '0' added to
// convert to ASCII, then 9 added because that's the bytecode command
// to access the thousands digit when the hundreds digit is odd, so
// the amount added is 10 + '0' + 9 = 57).
//
// LINENO_MID_TEMP is updated within the main loop, immediately after
// updating LINENO_MID, but because the bytecode interpreter reads
// from it it needs a valid value at the start of the loop.
.section .rodata
biascii_offset:
.byte 58, 59, 60, 61, 62, 63, 64, 65
.byte 66, 67, 68, 69, 70, 71, 72, 73
.byte 58, 59, 60, 61, 62, 63, 64, 65
.byte 66, 67, 68, 69, 70, 71, 72, 73
.text

// To save an instruction, precalculate minus the length of the
// bytecode. (Although the value of this is determined entirely by
// LINENO_WIDTH, the register it's stored in gets clobbered by
// system calls and thus needs to be recalculated each time.)
mov BYTECODE_NEG_LEN, BYTECODE_START
sub BYTECODE_NEG_LEN, BYTECODE_END

// The main loop

// The bytecode interpreter consists of four instructions:
// 1. Load the bytecode from memory into %ymm2;
// 2. Use it as a shuffle mask to shuffle LINENO_MID_TEMP;
// 3. Subtract the bytecode from the shuffle result;
// 4. Output the result of the subtraction.
//
// To see why this works, consider two cases. If the bytecode wants to
// output a literal character, then the shuffle will produce 0 for
// that byte (in AVX2, a shuffle with a a negative index produces an
// output of 0), and subtracting the bytecode from 0 then produces the
// character (because the bytecode encoded minus the character). If
// the bytecode instead wants to output a digit, then the shuffle will
// fetch the relevant digit from LINENO_MID_TEMP (which is the desired
// ASCII character plus the bytecode instruction that produces it),
// and subtract the bytecode instruction to just produce the character
// on its own.
//
// This produces an exactly correct line number as long as the line
// number approximation is within 100 of the true value: it will be
// correct as long as the relevant part of LINENO_MID is correct, and
// the worst case is for LINENO_MID to be storing, say, 200 and 300
// (the representation of 250) when the true line number is 400. The
// value in LINENO_MID specifically can be up to 50 away from the
// value of the line number as recorded by LINENO_MID and LINENO_LOW
// together, so as long as the line number registers are within 100,
// LINENO_MID will be within 150 (which is what is required).
//
// This doesn't update the bytecode instruction pointer or the pointer
// into the output buffer; those are updated once every 512 bytes (and
// to "advance the instruction pointer" the rest of the time, the main
// loop is unrolled, using hardcoded offsets with the pointer updates
// baked in).
//
// The bytecode instruction pointer itself is read from %rdx, not
// BYTECODE_IP, so that mid-loop arithmetic on BYTECODE_IP won't cause
// the interpreter to break.
//
// It's important to note one potential performance issue with this
// code: the read of the bytecode from memory is not only misalignable
// (vmovdqu); it splits a cache line 3/8 of the time. This causes L1
// split-load penalties on the 3/8 cycles where it occurs. I am not
// sure whether this actually reduces the program's performance in
// practice, or whether the split loads can be absorbed while waiting
// for writes to go through to the L2 cache. However, even if it does
// have a genuine performance cost, it seems like the least costly way
// to read the bytecode; structuring the bytecode to avoid split loads
// makes it take up substantially more memory, and the less cache that
// is used for the bytecode, the more that can be used for the output
// buffers. (In particular, increasing the bytecode to 2400 lines so
// that it's available at all four of the alignments required of it
// does not gain, because it then becomes so large that the processor
// struggles to keep it in L1 cache - it only just fits, and there
// isn't any way for it to know which parts of the cache are meant to
// stay in L1 and which are meant to leave to L2, so there's a large
// slowdown when it guesses wrong.)
#define INTERPRET_BYTECODE(bc_offset, buf_offset) \
vmovdqu %ymm2, [%rdx + bc_offset]; \
vpshufb %ymm0, LINENO_MID_TEMP, %ymm2; \
vpsubb %ymm0, %ymm0, %ymm2; \
vmovdqa [OUTPUT_PTR + buf_offset], %ymm0

// The main loop itself consists of sixteen uses of the bytecode
// interpreter, interleaved (to give the reorder buffer maximum
// flexibility) with all the other logic needed in the main loop.
// (Most modern processors can handle 4-6 instructions per clock cycle
// as long as they don't step on each others' toes; thus this loop's
// performance will be limited by the throughput of the L2 cache, with
// all the other work (bytecode interpretation, instruction decoding,
// miscellaneous other instructions, etc.) fitting into the gaps while
// the processor is waiting for the L2 cache to do its work.)

.p2align 5
main_loop:
// %rdx caches BYTECODE_IP's value at the start of the loop
mov %rdx, BYTECODE_IP
INTERPRET_BYTECODE(0, 0)

// %ymm1 caches LINENO_LOW's value at the start of the loop
vmovdqa %ymm1, LINENO_LOW
INTERPRET_BYTECODE(32, 32)

// Add LINENO_LOW_INCR to LINENO_LOW, checking for carry; it carried
// if the sign bit changed from 0 to 1. (vpandn is unintuitive; this
// is ~%ymm1 & LINENO_LOW, not %ymm1 & ~LINENO_LOW like the name
// suggests.)
INTERPRET_BYTECODE(64, 64)

vpandn %ymm3, %ymm1, LINENO_LOW
INTERPRET_BYTECODE(96, 96)

vpsrlq %ymm3, %ymm3, 63
INTERPRET_BYTECODE(128, 128)

// Add the carry to LINENO_MID (doubling it; LINENO_MID counts in
// units of 100 but a LINENO_LOW carry means 200).
INTERPRET_BYTECODE(160, 160)

INTERPRET_BYTECODE(192, 192)

vpmaxub LINENO_MID, LINENO_MID_BASE, LINENO_MID
INTERPRET_BYTECODE(224, 224)

// Update LINENO_MID_TEMP with the new value from LINENO_MID; this is
// the point at which the new value takes effect. This is done at the
// exact midpoint of the loop, in order to reduce the errors from
// updating once every 512 bytes as far as possible.
INTERPRET_BYTECODE(256, 256)

// Update the output and bytecode instruction pointers. The change to
// the output pointer kicks in immediately, but is cancelled out via
// the use of a negative offset until the end of the loop.
INTERPRET_BYTECODE(288, -224)

INTERPRET_BYTECODE(320, -192)

// The change to the bytecode instruction pointer doesn't kick in
// immediately, because it might need to wrap back to the start (this
// can be done by adding BYTECODE_NEG_LEN to it); this is why the
// interpreter has a cached copy of it in %rdx.
lea %rax, [BYTECODE_IP + BYTECODE_NEG_LEN]
INTERPRET_BYTECODE(352, -160)

INTERPRET_BYTECODE(384, -128)
// Some modern processors can optimise cmp better if it appears
// immediately before the command that uses the comparison result, so
// a couple of commands have been moved slightly to put the cmp next
// to the use of its result. With modern out-of-order processors,
// there is only a marginal advantage to manually interleaving the
// instructions being used, and the cmp advantage outweighs that.
cmp BYTECODE_IP, BYTECODE_END

cmovae BYTECODE_IP, %rax
INTERPRET_BYTECODE(416, -96)

INTERPRET_BYTECODE(448, -64)

INTERPRET_BYTECODE(480, -32)
cmp OUTPUT_PTR, OUTPUT_LIMIT
jb main_loop

after_main_loop:
// There are two reasons the main loop might terminate: either there's
// a pipeful of output, or the line number has increased in width
// (forcing the generaion of new bytecode to put more digits in the
// numbers being printed). In the latter case, a) the output may have
// overrun slightly, and OUTPUT_PTR needs to be moved back to
// OUTPUT_LIMIT:
mov OUTPUT_PTR, OUTPUT_LIMIT
// and b) there may be less than a pipeful of output, in which case it
// wouldn't be safe to output it and the swap_buffers call needs to be
// skipped. Calculate the pipe size into %rax, the amount of output
// into %rdx (swap_buffers needs it there anyway), and compare.
lea %rax, [%rip + iovec_base]
sub %rax, BYTECODE_START
neg %eax
mov %rdx, OUTPUT_PTR
and %edx, (2 << 20) - 1
cmp %edx, %eax
jb 1f
call swap_buffers

// If all the lines at this width have been exhausted, move to the
// next width.
1: test YMMS_AT_WIDTH, YMMS_AT_WIDTH
jnz check_lineno_top_carry

cmp LINENO_WIDTHe, 18  // third phase handles at most 18 digits
je fourth_phase

inc LINENO_WIDTHe
vpcmpeqb %ymm0, %ymm0, %ymm0
vpsubb ENDIAN_SHUFFLE, ENDIAN_SHUFFLE, %ymm0

lea GROUPS_OF_15, [GROUPS_OF_15 + GROUPS_OF_15 * 4]

jmp third_phase_per_width_init

// So far, the code has kept LINENO_MID and LINENO_LOW updated, but
// not LINENO_TOP. Because 10 billion lines of FizzBuzz don't normally
// have a length that's divisible by 512 (and indeed, vary in size a
// little because 10 billion isn't divisible by 15), it's possible for
// the 10-billions and higher digits to need to change in the middle
// of a main loop iteration - indeed, even in the middle of a single
// CPU instruction!
//
// It turns out that when discussing the line number registers above,
// I lied a little about the format. The bottom seven bytes of
// LINENO_MID do indeed represent the hundreds to hundred millions
// digits. However, the eighth changes in meaning over the course of
// the program. It does indeed represent the billions digit most of
// the time; but when the line number is getting close to a multiple
// of 10 billion, the billions and hundred-millions digits will always
// be the same as each other (either both 9s or both 0s). When this
// happens, the format changes: the hundred-millions digit of
// LINENO_MID represents *both* the hundred-millions and billions
// digits of the line number, and the top byte then represents the
// ten-billions digit. Because incrementing a number causes a row of
// consecutive 9s to either stay untouched, or all roll over to 0s at
// once, this effectively lets us do maths on more than 8 digits,
// meaning that the normal arithmetic code within the main loop can
// handle the ten-billions digit in addition to the digits below.
//
// Of course, the number printing code also needs to handle the new
// representation, but the number printing is done by a bytecode
// program, which can be made to output some of the digits being
// printed multiple times by repeating "print digit from LINENO_MID"
// commands within it. Those commands are generated from COUNTER_TOP
// anyway, so the program just changes the constant portion of
// COUNTER_TOP (and moves print-digit commands into the top half) in
// order to produce the appropriate bytecode changes.
//
// A similar method is used to handle carries in the hundred-billions,
// trillions, etc. digits.
//
// Incidentally, did you notice the apparent off-by-one in the
// initialisation of LINENO_MID within third_phase_per_width_init? It
// causes the "billions" digit to be initialised to 1 (not 0) when the
// line number width is 11 or higher. That's because the alternate
// representation will be in use during a line number width change (as
// higher powers of 10 are close to multiples of 10 billion), so the
// digit that's represented by that byte of LINENO_MID genuinely is a
// 1 rather than a 0.
check_lineno_top_carry:

// The condition to change line number format is:
// a) The line number is in normal format, and the hundred-millions
//    and billions digits are both 9; or
// b) The line number is in alternate format, and the hundred-millions
//    digit is 0.
// To avoid branchy code in the common case (when no format change is
// needed), REGEN_TRIGGER is used to store the specific values of the
// hundred-millions and billions digits that mean a change is needed,
// formatted as two repeats of billions, hundred-millions, 9, 9 in
// high-decimal (thus, when using normal format, REGEN_TRIGGER is
// high-decimal 99999999, i.e. -1 when interpreted as binary). The 9s
// are because vpshufd doesn't have very good resolution: the millions
// and ten-millions digits get read too, but can simply just be masked
// out. The two repeats are to ensure that both halves of LINENO_MID
// (the even-hundreds-digit and odd-hundreds-digit halves) have the
// correct value while changing (changing the format while half the
// register still ended ...98999999 would produce incorrect output).
vpshufd %xmm0, LINENO_MIDx, 0xED
vpextrq %rax, %xmm0, 0
mov %rdx, 0x0000ffff0000ffff
or %rax, %rdx
cmp %rax, REGEN_TRIGGER
jne calculate_main_loop_iterations

cmp REGEN_TRIGGER, -1
jne switch_to_normal_representation

switch_to_alternate_representation:
// Count the number of 9s at the end of LINENO_TOP. To fix an edge
// case, the top bit of LINENO_TOP is interpreted as a 0, preventing
// a 9 being recognised there (this causes 10**18-1 to increment to
// 10**17 rather than 10**18, but the program immediately exits
// before this can become a problem).
vpextrq %rdx, LINENO_TOPx, 1
mov SPILL, %rdx
shl %rdx, 1
shr %rdx, 1
not %rdx
bsf %rcx, %rdx
and %rcx, -8

// Change the format of LINENO_TOP so that the digit above the
// consecutive 9s becomes a reference to the top byte of LINENO_MID,
// and the 9s themselves references to the hundred-millions digit.
// This is done via a lookup table that specifies how to move the
// bytes around.
.section .rodata
alternate_representation_lookup_table:
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 7, 9, 10, 11, 12, 13, 14, 15
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 7, 9, 10, 11, 12, 13, 14, 15

.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 7, 10, 11, 12, 13, 14, 15
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 7, 10, 11, 12, 13, 14, 15

.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 7, 11, 12, 13, 14, 15
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 7, 11, 12, 13, 14, 15

.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 7, 12, 13, 14, 15
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 7, 12, 13, 14, 15

.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 7, 13, 14, 15
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 7, 13, 14, 15

.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 6, 7, 14, 15
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 6, 7, 14, 15

.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 6, 6, 7, 15
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 6, 6, 7, 15

.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 6, 6, 6, 7
.byte 0, 1, 2, 3, 4, 5, 6, 6
.byte 6, 6, 6, 6, 6, 6, 6, 7
.text

lea %rax, [%rip + alternate_representation_lookup_table]
vpshufb LINENO_TOP, LINENO_TOP, [%rax + 4 * %rcx]

// The top byte of LINENO_MID also needs the appropriate digit of
// LINENO_TOP placed there.
mov %rdx, SPILL
shr %rdx, %cl
vpinsrb LINENO_MIDx, LINENO_MIDx, %edx, 7
vpinsrb LINENO_MIDx, LINENO_MIDx, %edx, 15
vpermq LINENO_MID, LINENO_MID, 0x44

// Finally, REGEN_TRIGGER needs to store the pattern of digits that
// will prompt a shift back to the normal representation (the hundred-
// millions digit must be 0, and the value of the billions digit will
// be predictable).
inc %edx
shl %edx, 24
or %edx, 0xF6FFFF
mov REGEN_TRIGGERe, %edx
shl %rdx, 32
or REGEN_TRIGGER, %rdx
jmp generate_bytecode

switch_to_normal_representation:
// Switching back is fairly easy: LINENO_TOP can almost be converted
// back into its usual format by running the bytecode program stored
// there to remove any unusual references into LINENO_MID, then
// restoring the usual references manually. Running the program will
// unfortunately convert high-decimal to ASCII (or in this case zeroes
// because there's no need to do the subtraction), but that can be
// worked around by taking the bytewise maximum of the converted and
// original LINENO_TOP values (high-decimal is higher than bytecode
// references and much higher than zero).
vpsubb %ymm2, BASCII_OFFSET, LINENO_TOP
vpshufb %ymm0, LINENO_MID, %ymm2
vpmaxub LINENO_TOP, LINENO_TOP, %ymm0

// Manually fix the constant parts of lineno_top to contain their
// usual constant values
.section .rodata
lineno_top_max:
.byte 198, 197, 196, 195, 194, 193, 192, 191
.byte 255, 255, 255, 255, 255, 255, 255, 255
.byte 190, 189, 188, 187, 186, 185, 184, 183
.byte 255, 255, 255, 255, 255, 255, 255, 255
.text
vpminub LINENO_TOP, LINENO_TOP_MAX, LINENO_TOP

// The billions digit of LINENO_MID needs to be set back to 0 (which
// is its true value at this point: the same as the hundred-thousands
// digit, which is also 0).
vpsllq LINENO_MID, LINENO_MID, 8
vpsrlq LINENO_MID, LINENO_MID, 8
vpmaxub LINENO_MID, LINENO_MID_BASE, LINENO_MID

mov REGEN_TRIGGER, -1

jmp generate_bytecode

///// Fourth phase
//
// Ending at 999999999999999999 lines would be a little unsatisfying,
// so here's a routine to write the quintillionth line and exit.
//
// It's a "Buzz", which we can steal from the first phase's constant.

fourth_phase:

mov ARG1e, 1
lea ARG2, [%rip + fizzbuzz_intro + 11]
mov ARG3, 5
mov SYSCALL_NUMBER, __NR_write
syscall
call exit_on_error
xor ARG1e, ARG1e
jmp exit

///// Error handling code
//
// This doesn't run in a normal execution of the program, and isn't
// particularly optimised; I didn't comment it much because it isn't
// very interesting and also is fairly self-explanatory.

write_stderr:
mov ARG1e, 2
mov SYSCALL_NUMBER, __NR_write
syscall
ret

inefficiently_write_as_hex:
push %rax
push %rcx
shr %rax, %cl
and %rax, 0xF
.section .rodata
hexdigits: .ascii "0123456789ABCDEF"
.text
lea %rcx, [%rip + hexdigits]
movzx %rax, byte ptr [%rcx + %rax]
mov [%rip + error_write_buffer], %al
lea ARG2, [%rip + error_write_buffer]
mov ARG3e, 1
call write_stderr
pop %rcx
pop %rax
sub %ecx, 4
jns inefficiently_write_as_hex
ret

exit_on_error:
test SYSCALL_RETURN, SYSCALL_RETURN
js 1f
ret

.section .rodata
error_message_part_1: .ascii "Encountered OS error 0x"
error_message_part_2: .ascii " at RIP 0x"
error_message_part_3: .ascii ", exiting program.\n"
.text

1: push SYSCALL_RETURN
lea ARG2, [%rip + error_message_part_1]
mov ARG3e, 23
call write_stderr
pop SYSCALL_RETURN
neg SYSCALL_RETURN
mov %rcx, 8
call inefficiently_write_as_hex
lea ARG2, [%rip + error_message_part_2]
mov ARG3e, 10
call write_stderr
pop %rax  // find the caller's %rip from the stack
sub %rax, 5  // call exit_on_error compiles to 5 bytes
mov %rcx, 60
call inefficiently_write_as_hex
lea ARG2, [%rip + error_message_part_3]
mov ARG3e, 19
call write_stderr
mov ARG1e, 74
// fall through

exit:
mov SYSCALL_NUMBER, __NR_exit_group
syscall
ud2

.section .rodata
cpuid_error_message:
.ascii "Error: your CPUID command does not support command "
.ascii "0x80000006 (AMD-style L2 cache information).\n"
.text
lea ARG2, [%rip + cpuid_error_message]
mov ARG3e, 96
call write_stderr
mov ARG1e, 59
jmp exit

.section .rodata
pipe_error_message:
.ascii "This program can only output to a pipe "
.ascii "(try piping into cat?)\n"
.text
pipe_error:
lea ARG2, [%rip + pipe_error_message]
mov ARG3e, 64
call write_stderr
mov ARG1e, 73
jmp exit

.section .rodata
pipe_perm_error_message_part_1:
.ascii "Cannot allocate a sufficiently large kernel buffer.\n"
.ascii "Try setting /proc/sys/fs/pipe-max-size to 0x"
pipe_perm_error_message_part_2: .ascii ".\n"
.text
pipe_perm_error:
lea ARG2, [%rip + pipe_perm_error_message_part_1]
mov ARG3e, 96
call write_stderr
mov %rax, PIPE_SIZE
mov %ecx, 28
call inefficiently_write_as_hex
lea ARG2, [%rip + pipe_perm_error_message_part_2]
mov ARG3e, 2
call write_stderr
mov ARG1e, 77
jmp exit

.section .rodata
pipe_size_error_message_part_1:
.ascii "Failed to resize the kernel pipe buffer.\n"
.ascii "Requested size: 0x"
pipe_size_error_message_part_2: .ascii "\nActual size: 0x"
pipe_size_error_message_part_3:
.ascii "\n(If the buffer is too large, this may cause errors;"
.ascii "\nthe program could run too far ahead and overwrite"
.text
pipe_size_mismatch_error:
push SYSCALL_RETURN
lea ARG2, [%rip + pipe_size_error_message_part_1]
mov ARG3e, 59
call write_stderr
mov %rax, PIPE_SIZE
mov %ecx, 28
call inefficiently_write_as_hex
lea ARG2, [%rip + pipe_size_error_message_part_2]
mov ARG3e, 16
call write_stderr
pop %rax
mov %ecx, 28
call inefficiently_write_as_hex
lea ARG2, [%rip + pipe_size_error_message_part_3]
mov ARG3e, 141
call write_stderr
mov ARG1e, 73
jmp exit

• Wow, I have no words. This is really cool. Obviously didn't get a chance to understand much or any of it yet, but running it I'm getting 41-45GiB/s. This is almost an order of magnitude faster than most of the other submissions. Amazing work. Oct 26 '21 at 10:28
• Okay now that I actually bothered to read everything you wrote rather than hurrying up to try it out, I saw the taskset note. I'm getting 56GiB/s doing that! Oct 26 '21 at 10:38
• @BeeOnRope: I'm not totally sure, but I think what's happening is that the kernel's structures describing the pages from the process's memory that are being mapped into the pipe are being fought over by multiple CPUs; they're refcounted, and pv is busy trying to reduce the reference counts at the same time that fizzbuzz is trying to increase them. "Page table" was probably the wrong phrase, I was thinking about the kernel's paging structures rather than the processor's. Oct 29 '21 at 1:17
• @chx: I already have a master's thesis. This was harder. Oct 29 '21 at 1:17
• @StevenLu all Apple M1 chips are ARM chips and thus have a different instruction set to x86-64 with AVX2 extensions used here. Oct 29 '21 at 9:24

After much trial and error, with the goal of not resorting to Assembly while achieving the best single-threaded performance, this is my entry:

#include <unistd.h>

#define unlikely(e)      __builtin_expect((e), 0)
#define mcpy(d, s, n)    __builtin_memcpy((d), (s), (n))

#define CACHELINE 64
#define PGSIZE 4096
#define ALIGNED_BUF 65536
#define FIZZ "Fizz"
#define BUZZ "Buzz"
#define DELIM "\n"

typedef struct {
unsigned char offset;
char data[CACHELINE - sizeof(unsigned char)];
} counters_t;

static inline void os_write(int out, void *buf, unsigned int n)
{
while (n)
{
ssize_t written = write(out, buf, n);

if (written >= 0)
{
buf += written;
n -= written;
}
}
}

int main(void)
{
const int out = 1;

__attribute__((aligned(CACHELINE))) static counters_t counter = {
sizeof(counter.data) - 1, "00000000000000000000000000000000000000000000000000000000000000"
};
__attribute__((aligned(PGSIZE))) static char buf[ALIGNED_BUF + (sizeof(counter.data) * 15 * 3)] = { 0 };
char *off = buf;

for (;;)
{
// Write chunks of 30 counters until we reach ALIGNED_BUF
while (off - buf < ALIGNED_BUF)
{
#define NN (sizeof(counter.data) - 2)

// Hand-rolled counter copy, because with non-constant sizes the compiler
// just calls memcpy, which is too much overhead.
#define CTRCPY(i) do { \
const char *src = end; \
char *dst = off; \
unsigned _n = n; \
switch (_n & 3) { \
case 3: *dst++ = *src++; \
case 2: \
mcpy(dst, src, 2); \
dst += 2; src += 2; \
break; \
case 1: *dst++ = *src++; \
case 0: break; \
} \
for (_n &= ~3; _n; _n -= 4, dst += 4, src += 4) { \
mcpy(dst, src, 4); \
} \
mcpy(off + n, i DELIM, sizeof(i DELIM) - 1); \
off += n + sizeof(i DELIM) - 1; \
} while (0)

// Write the first 10 counters of the group (need to separate the
// first 10 counters from the rest of the chunk due to possible decimal
// order increase at the end of this block)
{
const char *const end = counter.data + counter.offset;
const unsigned int n = sizeof(counter.data) - counter.offset - 1;

CTRCPY("1"); // 1
CTRCPY("2"); // 2

mcpy(off, FIZZ DELIM, sizeof(FIZZ DELIM) - 1); // Fizz (3)
off += sizeof(FIZZ DELIM) - 1;

CTRCPY("4"); // 4

mcpy(off, BUZZ DELIM FIZZ DELIM, sizeof(BUZZ DELIM FIZZ DELIM) - 1); // Buzz (5) Fizz (6)
off += sizeof(BUZZ DELIM FIZZ DELIM) - 1;

CTRCPY("7"); // 7
CTRCPY("8"); // 8

mcpy(off, FIZZ DELIM BUZZ DELIM, sizeof(FIZZ DELIM BUZZ DELIM) - 1); // Fizz (9) Buzz (10)
off += sizeof(FIZZ DELIM BUZZ DELIM) - 1;

// Carry handling on MOD 10
for (unsigned d = NN; ; --d)
{
if (counter.data[d] != '9')
{
++counter.data[d];
break;
}
counter.data[d] = '0';
}

// Decimal order increases only when counter MOD 30 == 10
if (unlikely(counter.data[counter.offset - 1] != '0'))
{
if (unlikely(counter.offset == 1))
{
goto end;
}

--counter.offset;
}
}

// Write the chunk's remaining 20 counters
{
const char *const end = counter.data + counter.offset;
const unsigned int n = sizeof(counter.data) - counter.offset - 1;

CTRCPY("1"); // 11

mcpy(off, FIZZ DELIM, sizeof(FIZZ DELIM) - 1); // Fizz (12)
off += sizeof(FIZZ DELIM) - 1;

CTRCPY("3"); // 13
CTRCPY("4"); // 14

mcpy(off, FIZZ BUZZ DELIM, sizeof(FIZZ BUZZ DELIM) - 1); // FizzBuzz (15)
off += sizeof(FIZZ BUZZ DELIM) - 1;

CTRCPY("6"); // 16
CTRCPY("7"); // 17

mcpy(off, FIZZ DELIM, sizeof(FIZZ DELIM) - 1); // Fizz (18)
off += sizeof(FIZZ DELIM) - 1;

CTRCPY("9"); // 19

mcpy(off, BUZZ DELIM FIZZ DELIM, sizeof(BUZZ DELIM FIZZ DELIM) - 1); // Buzz (20) Fizz (21)
off += sizeof(BUZZ DELIM FIZZ DELIM) - 1;

// Carry handling on MOD 10
for (unsigned d = NN; ; --d)
{
if (counter.data[d] != '9')
{
++counter.data[d];
break;
}
counter.data[d] = '0';
}

CTRCPY("2"); // 22
CTRCPY("3"); // 23

mcpy(off, FIZZ DELIM BUZZ DELIM, sizeof(FIZZ DELIM BUZZ DELIM) - 1); // Fizz (24) Buzz (25)
off += sizeof(FIZZ DELIM BUZZ DELIM) - 1;

CTRCPY("6"); // 26

mcpy(off, FIZZ DELIM, sizeof(FIZZ DELIM) - 1); // Fizz (27)
off += sizeof(FIZZ DELIM) - 1;

CTRCPY("8"); // 28
CTRCPY("9"); // 29

mcpy(off, FIZZ BUZZ DELIM, sizeof(FIZZ BUZZ DELIM) - 1); // FizzBuzz (30)
off += sizeof(FIZZ BUZZ DELIM) - 1;

// Carry handling on MOD 10
for (unsigned d = NN; ; --d)
{
if (counter.data[d] != '9')
{
++counter.data[d];
break;
}
counter.data[d] = '0';
}
}
}

os_write(out, buf, ALIGNED_BUF);
mcpy(buf, buf + ALIGNED_BUF, (off - buf) % ALIGNED_BUF);
off -= ALIGNED_BUF;
}

end:
os_write(out, buf, off - buf);

return 0;
}


Compiled as clang -o fizz fizz.c -O3 -march=native (with clang 11.0.0 on my Ubuntu 20.10 installation, running kernel version 5.8.0-26.27-generic 5.8.14 on an Intel Core i7-8750H mobile CPU while plugged into the wall), this produces ~3.8GiB/s when run as ./fizz | pv > /dev/null (not very steady due to write blocking every once in a while, but there's nothing I can do about that when single-threaded, I guess).

EDIT: Optimised the carry handling a bit, and now I'm getting ~3.9GiB/s on my machine (same configuration as above).

• Welcome to the site! Nice first post. Nov 14 '20 at 18:38
• Interesting, GCC sees through your memcpy(dst,src,4) loop and decides to turn into one variable-sized call to libc memcpy. godbolt.org/z/14K5Pd. But clang does turn that nasty-looking memcpy(dst,src,4) repeat-loop into vmovups instructions that load/store 32 bytes at a time, not just 4! godbolt.org/z/jM3f8K. So basically I think it turns (part of?) your manual memcpy back into one big memcpy like gcc does, but then inlines it instead of calling the libc function. Nov 16 '20 at 20:23
• You don't need a macro for __builtin_memcpy, just use memcpy. GCC and clang define memcpy as a builtin (same as __builtin_memcpy) by default, unless you use -fno-builtin-memcpy or -fno-builtin (which you'd use for writing unit-tests or benchmarks for a hand-written libc or kernel implementation for example, to make it actually call the function instead of inlining or optimizing away). Nov 16 '20 at 20:26
• @PeterCordes you're right about __builtin_memcpy in this case, but as an embedded developer (where GCC/clang don't assume that memcpy can be inlined with constant size) it's an instinct for me to reach for __builtin_memcpy in order to ensure inlining. Nov 16 '20 at 20:37
• Ok, so you're used to compiling with -fno-builtin for embedded systems. Oh right, that's implied by -ffreestanding, unfortunately even for the functions gcc requires the freestanding environment to provide (memcpy, memmove, memset and memcmp). That's unfortunate; you end up wanting a #define to the __builtin version for every ISO C standard function you do implement. Nov 16 '20 at 20:49

I was struggling to get more than 2.75GB/s on my rig but then I realised I wasn't compiling with -O3 which bumped me up to 6.75GB/s.

#include <stdio.h>
#include <string.h>
#include <unistd.h>
char buf[416];
char out[65536 + 4096] = "1\n2\nFizz\n4\nBuzz\nFizz\n7\n8\nFizz\n";
int main(int argc, char **argv) {
const int o[16] = { 4, 7, 2, 11, 2, 7, 12, 2, 12, 7, 2, 11, 2, 7, 12, 2 };
char *t = out + 30;
unsigned long long i = 1, j = 1;
for (int l = 1; l < 20; l++) {
int n = sprintf(buf, "Buzz\n%llu1\nFizz\n%llu3\n%llu4\nFizzBuzz\n%llu6\n%llu7\nFizz\n%llu9\nBuzz\nFizz\n%llu2\n%llu3\nFizz\nBuzz\n%llu6\nFizz\n%llu8\n%llu9\nFizzBuzz\n%llu1\n%llu2\nFizz\n%llu4\nBuzz\nFizz\n%llu7\n%llu8\nFizz\n", i, i, i, i, i, i, i + 1, i + 1, i + 1, i + 1, i + 1, i + 2, i + 2, i + 2, i + 2, i + 2);
i *= 10;
while (j < i) {
memcpy(t, buf, n);
t += n;
if (t >= &out[65536]) {
char *u = out;
do {
int w = write(1, u, &out[65536] - u);
if (w > 0) u += w;
} while (u < &out[65536]);
memcpy(out, out + 65536, t - &out[65536]);
t -= 65536;
}
char *q = buf;
for (int k = 0; k < 16; k++) {
char *p = q += o[k] + l;
if (*p < '7') *p += 3;
else {
*p-- -= 7;
while (*p == '9') *p-- = '0';
++*p;
}
}
j += 3;
}
}
}

• Nicely done! The printed buffer reuse is something that intended to do as well, but didn't get around to due to the complexity of my code. Impressive work! Nov 15 '20 at 14:46
• Very nice. I tried parallelizing your and @IsaacG.'s solutions, but unfortunately two threads is already enough to cause enough contention on my 3900X to get less overall performance out of it than out of the single-threaded versions! Two threads pinned to the right cores seem to get it up to 8 GB/s for just a second or two, probably just the CPU boosting frequencies for a moment. Afterwards there is no benefit. Nov 15 '20 at 18:56
• What CPU (model number and/or uarch and frequency it actually ran at), compiler version, and OS? Benchmark numbers are most useful when we know the test system. Nov 16 '20 at 18:41
• @PeterCordes My rig? it's an AMD Ryzen 7 3700X 8-core @ 2.2-3.6GHz using GCC 9.3.1 on Fedora 31.
– Neil
Nov 16 '20 at 18:50
• Consider using -O3 -march=native to let it take advantage of CPU features like AVX (32-byte vectors) if it ever wants to inline a memcpy or something. (The -mtune=znver2 implied by -march on your machine may help, too.) With memcpy size being runtime variable, it will just call the library function. so -fno-plt should reduce overhead for each call. Also -fno-pie -no-pie may make global arrays slightly more efficient (mov-immediate instead of RIP-relative LEA to put the address in a register, or t-&out[65536] could be a sub-immediate or maybe even LEA.) Nov 16 '20 at 20:10

I tweaked Neil's code a bit (so most credit goes to him) and managed to squeeze some more performance out of it; I also prepared it for unrolling more loops but ultimately I gave up (that's why the code is unreadable gobbledygook).

#include <stdio.h>
#include <string.h>
#include <unistd.h>

#define f(Z) {char*p=q+=Z+l;if(*p<'7')*p+=3;else{*p---=7;while(*p=='9')*p--='0';++*p;}}
#define v(N) {while(j<i){memcpy(t,buf,N);t+=n;if(t>=&out[65536]){char*u=out; \
do{int w=write(1,u,&out[65536]-u);if(w>0)u+=w;}while(u<&out[65536 \
]);memcpy(out,out+65536,t-&out[65536]);t-=65536;}char*q=buf;f(4); \
f(7);f(2);f(11);f(2);f(7);f(12);f(2);f(12);f(7);f(2);f(11);f(2);f \
(7);f(12);f(2);j+=3;}}
char buf[256];
char out[65536 + 4096] = "1\n2\nFizz\n4\nBuzz\nFizz\n7\n8\nFizz\n";
int main(void) {
char *t = out + 30;
unsigned long long i = 1, j = 1;
for (int l = 1; l < 20; l++) {
int n=sprintf(buf, "Buzz\n%llu1\nFizz\n%llu3\n%llu4\nFizzBuzz\n%llu6\n%llu7\nFizz\n%llu9\nBuzz\nFizz\n%llu2\n%llu3\nFizz\nBuzz\n%llu6\nFizz\n%llu8\n%llu9\nFizzBuzz\n%llu1\n%llu2\nFizz\n%llu4\nBuzz\nFizz\n%llu7\n%llu8\nFizz\n", i, i, i, i, i, i, i + 1, i + 1, i + 1, i + 1, i + 1, i + 2, i + 2, i + 2, i + 2, i + 2);
i*=10;
v(n);
}
return 0;
}


On my PC, Neil's submission is ~5% slower. I also tried it on friend's Intel box and the tweaked version is faster.

• Added to the scoreboard. This is the best one so far. You seem to also have made Neil's submission much more "stable" (less variance in the output rate). Nov 15 '20 at 22:59
• @IsaacG pointed out that my buf is actually too small and my program would crash if it ever got to the really high numbers.
– Neil
Nov 16 '20 at 22:32
• @Neil Yeah, I noticed this too, but if I tried to enlargen it the performance gets a solid drop. So we're just hoping it works lol Nov 17 '20 at 6:25
• Huh, so it does, but I tried 416 and it seems to be OK, so...
– Neil
Nov 17 '20 at 10:09
• i feel like the buffer is too big to fit in l1 cache, not that it affects anything... Feb 9 '21 at 6:41

Here is my attempt at using just-in-time compilation to emit fast FizzBuzz assembly that is specialized for every digit length. It's basically the same idea as Neil's answer, just more overengineered. A further 2x comes from the vmsplice system call as in the winning answer. While there are a few other similarities in the AVX2 code, the usage of vmsplice is the only bit that I downright "stole" from there; all the vector code is my own.

The basic idea is to extract 32 bytes out of a prebuilt set of 32 characters that includes the current line number divided by ten (lo_bytes), the digits 0-9 (hi_bytes 0-9), the letters in Fizz and Buzz (hi_bytes 11-15) and the newline character (hi_bytes byte 10). There are some extra complications:

• about half of the time the 32 bytes must be extracted in two steps, with the increment of lo_bytes inserted between the two extractions. The alternation of "mov", "or", "store", "increment lo_bytes" and "end of run" operations is stored as a kind of "bytecode". First, the program generates a string with the template of the FizzBuzz output for 30 consecutive numbers (30 is the LCM of 3, 5 and 10); then, to generate the bytecode, it operates on three substrings corresponding to 10 consecutive numbers.

• the AVX2 vpshufb instruction operates on two "lanes" of 128 bits. Therefore it can only gather from bytes 0-15 into bytes 0-15, and from bytes 16-31 into bytes 16-31. It can also place a zero in any byte though, which comes in really handy. This is why there are separate "lo_bytes" and "hi_bytes". Each mask is split in two parts, one for the "lo_bytes" and one for the "hi_bytes"; each is vpshufb'ed while filling the bytes that come from the "wrong" character with a zero, and then the two parts are ORed together.

I mentioned the bytecode before. There are two reasons why the program does not go directly to x86 code. First, going through bytecode simplifies noticeably the JIT compiler. Second, the JIT-compiled inner loop does not handle carry from byte 7 to byte 8 of the lo_bytes (which happens every 10^9 numbers written); that part is handled by interpreting the bytecode.

Actually, there's a third reason to have the bytecode, and it is possibly the most important even though it doesn't apply to the code submitted below. The bytecode approach separates very well the tasks of preprocessing (figuring out the exact sequence for each number length) and emitting output; therefore, during development I could first work on the preprocessor while keeping a stupid for loop for the output, then add vectorized C code (which actually survives in the slow path to handle carries), and finally generated the code on the fly. Of course the program was super slow until the introduction of the JIT, but being able to test AVX2 code in C is obviously much easier! The whole implementation only took about 6 hours, hence more optimization is probably possible (sizing the piping buffer, better scheduling of the x86 code, etc.).

The hints about taskset and the same warnings about needing a "useless cat" apply to this program as well due to the use of vmsplice.

The code is not super polished. Variable names are especially horrible, sorry about that.

/*
* Author: Paolo Bonzini
* gcc fb.c -o fb -O2 -g -mavx -mavx2 -flax-vector-conversions
*/
#define _GNU_SOURCE

#include <sys/mman.h>
#include <fcntl.h>
#include <sys/uio.h>
#include <unistd.h>
#include <assert.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <immintrin.h>
#include <avxintrin.h>

#define F_SETPIPE_SZ 1031

#define ZERO    15

typedef uint8_t v32qi __attribute__((vector_size(32), aligned(32)));
typedef uint8_t v32qi_u __attribute__((vector_size(32), aligned(1)));

v32qi lo_bytes = {
'1', '0', '0', '0', '0', '0', '0', '0',     /* 0 */
'0', '0', '0', '0', '0', '0', '0', '\0',    /* 8 */
'1', '0', '0', '0', '0', '0', '0', '0',     /* 0 */
'0', '0', '0', '0', '0', '0', '0', '\0',    /* 8 */
};

uint8_t hi_bytes[16] = {
'0', '1', '2', '3', '4', '5', '6', '7',     /* 16 */
'8', '9', '\n', 'z', 'u', 'B', 'i', 'F',    /* 24 */
};

static v32qi biased_zero = {
246, 246, 246, 246, 246, 246, 246, 246,
246, 246, 246, 246, 246, 246, 246, 246,
246, 246, 246, 246, 246, 246, 246, 246,
246, 246, 246, 246, 246, 246, 246, 246,
};

static v32qi biased_line = {
247, 246, 246, 246, 246, 246, 246, 246,
246, 246, 246, 246, 246, 246, 246, 246,
247, 246, 246, 246, 246, 246, 246, 246,
246, 246, 246, 246, 246, 246, 246, 246,
};

1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0,
};

0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 0, 0,
};

#define OUTPUT_SIZE(n)      (94+16*n)
#define TEMPLATE_SIZE(n)    ((OUTPUT_SIZE(n) + 31) & ~31)
#define MAX         15

#define OUTBUF_SIZE     1048576

static uint8_t template[TEMPLATE_SIZE(MAX)];
static uint8_t *output1;
static uint8_t *output2;
static uint8_t *output;
#define BOUNDARY (output + OUTBUF_SIZE)

static uint8_t code_buffer[64];
static uint8_t *code_ptr;

static uint8_t *jit_buffer;
static uint8_t *jit_ptr;
typedef uint8_t *jit_fn(uint8_t *, int);

/*
* Bytecode language:
*   0 = mov 32 bytes into temp buffer from the next mask
*   1 = or 32 bytes into temp buffer from the next mask
*   2 = increment the line
*   3 = store 32 bytes from temp buffer
*   -32..-1 = add n to the output pointer
*/
static void gen_prolog(void)
{
code_ptr = code_buffer;
}

static void gen_epilog(int template_size)
{
*code_ptr++ = 3;
*code_ptr++ = (template_size & 31) - 32;
}

static void do_gen_or_code(int from)
{

// o[i++] |= out_bytes[template[from + i]];
for (int i = 0; i < 32; i++) {
uint8_t m = template[from + i];
if (m < 16) {
} else {
}
}
*code_ptr++ = 1;
}

static void do_gen_mov_code(int from, int to)
{

// o[i++] = out_bytes[template[from + i]];
for (int i = 0; i < 32; i++) {
uint8_t m = (from + i > to) ? ZERO : template[from + i];
if (m < 16) {
} else {
}
}
*code_ptr++ = 0;
}

static void gen_inc_code(void)
{
*code_ptr++ = 2;
}

static void gen_out_code(int from, int to)
{
int offset = from & ~31;
if (offset < from) {
assert(to >= offset + 32);
do_gen_or_code(offset);
offset += 32;
*code_ptr++ = 3;
}
while (offset < to) {
do_gen_mov_code(offset, to);
offset += 32;
if (offset <= to)
*code_ptr++ = 3;
}
memset(template + from, ZERO, to - from);
}

{
v32qi old = biased_line;
biased_line = _mm256_max_epu8(incr, biased_zero);
lo_bytes += biased_line - old;
}

{
}

static uint8_t *maybe_output(uint8_t *o)
{
if (o > output + OUTBUF_SIZE) {
#if 1
struct iovec iov = {output, OUTBUF_SIZE};
do {
ssize_t r = vmsplice(1, &iov, 1, 0);
if (r < 0) {
perror("vmsplice");
exit(1);
}
iov.iov_base += r;
iov.iov_len -= r;
} while (iov.iov_len);
#else
write(1, output, OUTBUF_SIZE);
#endif
if (output == output1) {
memcpy(output2, BOUNDARY, o - BOUNDARY);
o = output2 + (o - BOUNDARY);
output = output2;
} else {
memcpy(output1, BOUNDARY, o - BOUNDARY);
o = output1 + (o - BOUNDARY);
output = output1;
}
}
return o;
}

static uint8_t *slow_run(uint8_t *o, int carry)
{
const uint8_t *p;
v32qi temp;
for (p = code_buffer; p < code_ptr; p++) {
uint8_t c = *p;
if (c == 0) {
temp = do_shuffle(m);
m += 2;
} else if (c == 1) {
temp |= do_shuffle(m);
m += 2;
} else if (c == 3) {
*(v32qi_u *)o = temp;
o += 32;
} else if (c == 2) {
if (--carry == 0)
} else {
o += (int8_t) c;
}
}
return maybe_output(o);
}

#define o(b) (*jit_ptr++ = (0x##b))
#define s(p) jit_ptr += ({ uint32_t x = (uintptr_t)p - (uintptr_t)(jit_ptr + 4); memcpy(jit_ptr, &x, 4); 4; })
#define d(i) jit_ptr += ({ uint32_t x = (i); memcpy(jit_ptr, &x, 4); 4; })

void compile(void)
{
const uint8_t *p, *label;
int ofs = 0;

jit_ptr = jit_buffer;
o(C5),o(FD),o(6F),o(05),s(&lo_bytes);      // vmovdqa ymm0, lo_bytes
o(C5),o(FD),o(6F),o(15),s(&biased_line);   // vmovdqa ymm2, biased_line
o(C5),o(FD),o(6F),o(1D),s(&biased_zero);   // vmovdqa ymm3, biased_zero

/* in inc_line, lo_bytes - old is always the same.  Put it in ymm1.  */
o(C5),o(FD),o(F8),o(CA);           // vpsubb ymm1, ymm0, ymm2

label = jit_ptr;
for (p = code_buffer; p < code_ptr; p++) {
uint8_t c = *p;
if (c == 0) {
m++;
m++;
o(C4),o(E2),o(7D),o(00),o(F6); // vpshufb ymm6, ymm0, ymm6
o(C5),o(D5),o(EB),o(EE);       // vpor ymm5, ymm5, ymm6
} else if (c == 1) {
m++;
m++;
o(C4),o(E2),o(7D),o(00),o(F6); // vpshufb ymm6, ymm0, ymm6
o(C5),o(D5),o(EB),o(EF);       // vpor ymm5, ymm5, ymm7
o(C5),o(D5),o(EB),o(EE);       // vpor ymm5, ymm5, ymm6
} else if (c == 3) {
o(C5),o(FE),o(7F),o(AF),d(ofs); // vmovdqu [rdi+NNN], ymm5
ofs += 32;
} else if (c == 2) {
o(C5),o(ED),o(D4),o(D4);      // vpaddq ymm2, ymm2, ymm4
o(C5),o(ED),o(DE),o(D3);      // vpmaxub ymm2, ymm2, ymm3
o(C5),o(F5),o(FC),o(C2);      // vpaddb ymm0, ymm1, ymm2
} else {
ofs += (int8_t) c;
}
}
o(FF),o(CE);                // dec esi
o(0F),o(85),s(label);       // jnz label
o(48),o(89),o(F8);          // mov rax, rdi
o(C5),o(FD),o(7F),o(05),s(&lo_bytes);     // vmovdqa lo_bytes, ymm0
o(C5),o(FD),o(7F),o(15),s(&biased_line);  // vmovdqa biased_line, ymm2
o(C3);                // ret
}

#define INITIAL "1\n2\nFizz\n4\nBuzz\nFizz\n7\n8\nFizz\n"

int main()
{
uint8_t shuffle[] = { 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 16, 26 };
const uint8_t fizz[] = { 31, 30, 27, 27, 26 };
const uint8_t fizzbuzz[] = { 31, 30, 27, 27, 29, 28, 27, 27, 26 };
const uint8_t *buzz = fizzbuzz + 4;

int l;
uint64_t n;
uint32_t tens_till_carry = TENS_FOR_VPADDQ - 1;

output1 = mmap(NULL, OUTBUF_SIZE + 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANON, -1, 0);
output2 = mmap(NULL, OUTBUF_SIZE + 4096, PROT_READ|PROT_WRITE, MAP_PRIVATE|MAP_ANON, -1, 0);
output = output1;
uint8_t *o = mempcpy(output, INITIAL, strlen(INITIAL));

fcntl(1, F_SETPIPE_SZ, OUTBUF_SIZE);
memset(template, ZERO, sizeof(template));

MAP_32BIT|MAP_PRIVATE|MAP_ANON, -1, 0);
assert((uintptr_t)jit_buffer <= 0x7FFFFFFF);

for (l = 2, n = 3; l <= MAX; l++, n = n * 10) {
int output_size = OUTPUT_SIZE(l);
int template_size = TEMPLATE_SIZE(l);

uint8_t *s = shuffle + sizeof(shuffle) - l - 1;
uint8_t *p = template;

#define ZERO_UNITS s[l - 1] = 16;
#define INC_UNITS s[l - 1]++;

ZERO_UNITS; p = mempcpy(p, buzz, 5); // 10
INC_UNITS; p = mempcpy(p, s, l + 1); // 11
INC_UNITS; p = mempcpy(p, fizz, 5); // 12
INC_UNITS; p = mempcpy(p, s, l + 1); // 13
INC_UNITS; p = mempcpy(p, s, l + 1); // 14
INC_UNITS; p = mempcpy(p, fizzbuzz, 9); // 15
INC_UNITS; p = mempcpy(p, s, l + 1); // 16
INC_UNITS; p = mempcpy(p, s, l + 1); // 17
INC_UNITS; p = mempcpy(p, fizz, 5); // 18
INC_UNITS; p = mempcpy(p, s, l + 1); // 19
ZERO_UNITS; p = mempcpy(p, buzz, 5); // 20
INC_UNITS; p = mempcpy(p, fizz, 5); // 21
INC_UNITS; p = mempcpy(p, s, l + 1); // 22
INC_UNITS; p = mempcpy(p, s, l + 1); // 23
INC_UNITS; p = mempcpy(p, fizz, 5); // 24
INC_UNITS; p = mempcpy(p, buzz, 5); // 25
INC_UNITS; p = mempcpy(p, s, l + 1); // 26
INC_UNITS; p = mempcpy(p, fizz, 5); // 27
INC_UNITS; p = mempcpy(p, s, l + 1); // 28
INC_UNITS; p = mempcpy(p, s, l + 1); // 29
ZERO_UNITS; p = mempcpy(p, fizzbuzz, 9); // 30
INC_UNITS; p = mempcpy(p, s, l + 1); // 31
INC_UNITS; p = mempcpy(p, s, l + 1); // 32
INC_UNITS; p = mempcpy(p, fizz, 5); // 33
INC_UNITS; p = mempcpy(p, s, l + 1); // 34
INC_UNITS; p = mempcpy(p, buzz, 5); // 35
INC_UNITS; p = mempcpy(p, fizz, 5); // 36
INC_UNITS; p = mempcpy(p, s, l + 1); // 37
INC_UNITS; p = mempcpy(p, s, l + 1); // 38
INC_UNITS; p = mempcpy(p, fizz, 5); // 39
memset(p, ZERO, template + template_size - p);

gen_prolog();
gen_out_code(0, 30+6*l);
gen_inc_code();
gen_out_code(30+6*l, 60+11*l);
gen_inc_code();
gen_out_code(60+11*l, 94+16*l);
gen_inc_code();
gen_epilog(94+16*l);

compile();

uint64_t left = n;
do {
int runs;
if (tens_till_carry <= 3) {
if (tens_till_carry == 0) {
runs = 0;
} else {
o = slow_run(o, tens_till_carry);
runs = 1;
}
} else {
runs = (BOUNDARY - o) / output_size + 1;
if (runs > left)
runs = left;
if (runs * 3 > tens_till_carry)
runs = tens_till_carry / 3;

o = ((jit_fn *)jit_buffer) (o, runs);
o = maybe_output(o);
}
left -= runs;
tens_till_carry -= runs * 3;
} while (left);
}
write(1, output, o - output);
}


Last minute add: here is how the number of source code lines grew as the bells and whistles were added. Handling the pesky carry is more work than the JIT compiler!

227 basic implementation
252 rewrite mask operations as AVX2
258 move loop inside the run function
276 increment line number using AVX
332 handle carry over 10^8
377 generate custom AVX2 code for each length
404 use the "vmsplice" system call

• Welcome to Code Golf, and fantastic first answer! I don't know much about this, but I'm curious to see how fast it ends up being. Oct 29 '21 at 16:51
• Stable 23.3GiB/s! had to compile with gcc -mavx2 -flax-vector-conversions paolo.c, is that alright? Oct 29 '21 at 17:39
• Yet again, I miss(understood) the taskset note. 40GiB/s with taskset Oct 29 '21 at 17:52
• And I also missed you specified the exact compilation command in a comment at the top of the code. Tried it now with your specific flags - it made no difference compared to my previous compilation command - still hangs around ~40-41GiB/s Oct 29 '21 at 17:59
• I suspect their calculation of the L2 size is not optimal on my computer, and my choice of the constant 1048576 might not be the best on yours. On my laptop my program is actually faster, so I had hopes (!), but the winning answer is awesome and they deserve first place. I'm happy I could get so close with my relatively simple algorithm, that I could implement in my spare time over less than a week. Oct 29 '21 at 22:18

Coded in rust- modern languages can be fast too. Build with cargo build --release* and run with ./target/release/fizz_buzz. The count goes up by 15 every iteration of the loop. The itoap crate is used to quickly write integers to the buffer. Adds 15 line chunks to an array unless there isn't enough space left in the buffer for a max-sized chunk, and when that happens it flushes the buffer to stdout.

main.rs:

use std::io::*;
use itoap::Integer;
const FIZZ:*const u8 = "Fizz\n".as_ptr();
const BUZZ:*const u8 = "Buzz\n".as_ptr();
const FIZZBUZZ:*const u8 = "FizzBuzz\n".as_ptr();
const BUF_SIZE:usize = 1024*256;
const BLOCK_SIZE:usize = 15 * i32::MAX_LEN;
/// buf.len() > count
macro_rules! itoap_write{
($buf:ident,$count:ident,$num:ident)=>{$count += itoap::write_to_ptr(
$buf.get_unchecked_mut($count..).as_mut_ptr(),
$num );$buf.as_mut_ptr().add($count).write(b'\n');$count += 1;
}
}
///ptr must be valid, buf.len() > count, ptr.add(len) must not overflow buffer
macro_rules! str_write{
($buf:ident,$count:ident,$ptr:ident,$len:literal)=>{
let ptr = $buf.get_unchecked_mut($count..).as_mut_ptr();
ptr.copy_from_nonoverlapping($ptr,$len);
$count +=$len;
}
}

fn main() -> Result<()>{
let mut write = stdout();
let mut count:usize = 0;
let mut buf = [0u8;BUF_SIZE];
let mut i:i32 = -1;
loop{
if &count + &BLOCK_SIZE > BUF_SIZE{
unsafe{
write.write_all(
buf.get_unchecked(..count)
)?;
}
count = 0;
}
i += 2;
unsafe{
itoap_write!(buf,count,i);
i += 1;
itoap_write!(buf,count,i);
str_write!(buf,count,FIZZ,5);
i += 2;
itoap_write!(buf,count,i);
str_write!(buf,count,BUZZ,5);
str_write!(buf,count,FIZZ,5);
i += 3;
itoap_write!(buf,count,i);
i += 1;
itoap_write!(buf,count,i);
str_write!(buf,count,FIZZ,5);
str_write!(buf,count,BUZZ,5);
i += 3;
itoap_write!(buf,count,i);
str_write!(buf,count,FIZZ,5);
i += 2;
itoap_write!(buf,count,i);
i += 1;
itoap_write!(buf,count,i);
str_write!(buf,count,FIZZBUZZ,9);
}
}
}



Cargo.toml:

[package]
name = "fizz_buzz"
version = "0.1.0"
authors = ["aiden4"]
edition = "2018"

[dependencies]
itoap = "0.1"
[[bin]]
name = "fizz_buzz"
path = "main.rs"

[profile.release]
lto = "fat"


*requires cargo to be able to connect to the internet

My code works on Windows 10. It outputs 8-9 GiB/s when the CPU is cool enough.

I used the following ideas in my code:

• Filling a buffer 256 KiB and sending it to output; for smaller buffer size the performance suffers; bigger buffer sometimes improves performance, but never by much.
• For numbers which have the same number of digits, it works in chunks of 15 output lines. These chunks have identical length. While the size of the output buffer is big enough, it copies the previous chunk and adds 15 to the ASCII representation of all the numbers in it.
• Near the end of the buffer and for first chunk, it calculates the output messages explicitly. Also, if numbers in the chunk have different length (e.g. 9999 and 10000).
• It uses OpenMP to calculate 4 chunks simultaneously. I set NUM_THREADS to 4 (best on my computer, which has 8 logical cores); a larger setting might be better.

When I want to verify the output, I set check_file = 1 in code; if check_file = 0, it writes to NUL, which is the null output device on Windows.

#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <inttypes.h>
#include <assert.h>
#include "windows.h"
#include "fileapi.h"
#include "process.h"

int size_15(int num_digits) // size of 15 messages, where numbers have a given number of digits
{
return 8 * num_digits + 47;
}

int num_of_digits(int64_t counter) // number of digits
{
int result = 1;
if (counter >= 100000000)
{
counter /= 100000000;
result += 8;
}
if (counter >= 10000)
{
counter /= 10000;
result += 4;
}
if (counter >= 100)
{
counter /= 100;
result += 2;
}
if (counter >= 10)
return result + 1;
else
return result;
}

void print_num(char* buf, int64_t counter, int num_digits)
{
for (int i = 0; i < num_digits; ++i)
{
buf[num_digits - 1 - i] = counter % 10 + '0';
counter /= 10;
}
}

{
char digit = p[num_digits - 1] + 5;
int c = (digit > '9');
p[num_digits - 1] = (char)(digit - c * 10);
c += 1;
for (int i = 1; i < num_digits; ++i)
{
if (c == 0)
break;
digit = (char)(p[num_digits - 1 - i] + c);
c = digit > '9';
p[num_digits - 1 - i] = (char)(digit - c * 10);
}
}

uint64_t fill_general(char* buf, int size, uint64_t counter, int* excess)
{
char* p = buf;
while (p < buf + size)
{
int fizz = counter % 3 == 0;
int buzz = counter % 5 == 0;
if (fizz && buzz)
{
memcpy(p, "FizzBuzz\n", 9);
p += 9;
}
else if (fizz)
{
memcpy(p, "Fizz\n", 5);
p += 5;
}
else if (buzz)
{
memcpy(p, "Buzz\n", 5);
p += 5;
}
else
{
int num_digits = num_of_digits(counter);
print_num(p, counter, num_digits);
p[num_digits] = '\n';
p += num_digits + 1;
}
++counter;
}
*excess = (int)(p - (buf + size));
return counter;
}

void fill15(char* buf, int64_t counter, int num_digits, int num_ofs[8])
{
char* p = buf;
int m15 = counter % 15;
for (int i = m15; i < m15 + 15; ++i)
{
if (i % 15 == 0)
{
memcpy(p, "FizzBuzz\n", 9);
p += 9;
}
else if (i % 3 == 0)
{
memcpy(p, "Fizz\n", 5);
p += 5;
}
else if (i % 5 == 0)
{
memcpy(p, "Buzz\n", 5);
p += 5;
}
else
{
*num_ofs++ = (int)(p - buf);
print_num(p, counter + i - m15, num_digits);
p += num_digits;
*p++ = '\n';
}
}
}

// memcpy replacement; works only for sizes equal to 47 + 8 * n, for small n
void copy_47_8n(char* src, unsigned size)
{
char* dst = src + size;

memcpy(dst, src, 47);
size -= 47;
dst += 47;
src += 47;

if (size >= 128)
exit(1);
if (size >= 96)
memcpy(dst + 64, src + 64, 32);
if (size >= 64)
memcpy(dst + 32, src + 32, 32);
if (size >= 32)
memcpy(dst + 0, src + 0, 32);
dst += size / 32 * 32;
src += size / 32 * 32;
size %= 32;
if (size >= 24)
memcpy(dst + 16, src + 16, 8);
if (size >= 16)
memcpy(dst + 8, src + 8, 8);
if (size >= 8)
memcpy(dst + 0, src + 0, 8);
}

uint64_t fill_fast(char* buf, int size, uint64_t counter, int* excess)
{
const int num_digits = num_of_digits(counter);
const int chunk_size = 8 * num_digits + 47;
const int num_iter = size / chunk_size;
#pragma omp parallel for
{
char* output = buf + begin_iter * chunk_size;
int num_ofs[8];
fill15(output, counter + begin_iter, num_digits, num_ofs);
for (int iter = 1; iter < thread_num_iter; ++iter)
{
copy_47_8n(output, chunk_size);
for (int i = 0; i < 8; ++i)
add_15_to_decimal_num(output + chunk_size + num_ofs[i], num_digits);
output += chunk_size;
}
}

buf += num_iter * chunk_size;
size -= num_iter * chunk_size;
counter += num_iter * 15;

return fill_general(buf, size, counter, excess);
}

uint64_t fill(char* buf, int size, uint64_t counter, int* excess)
{
int num_digits = num_of_digits(counter);
int64_t max_next_counter = counter + size / (8 * num_digits + 47) * 15 + 15;
int max_next_num_digits = num_of_digits(max_next_counter);
if (num_digits == max_next_num_digits)
return fill_fast(buf, size, counter, excess);
else
return fill_general(buf, size, counter, excess);
}

void file_io(void)
{
int check_file = 0;
HANDLE f = CreateFileA(check_file ? "my.txt" : "NUL", GENERIC_WRITE, FILE_SHARE_READ, 0, CREATE_ALWAYS, FILE_ATTRIBUTE_NORMAL, 0);
DWORD e = GetLastError();
LARGE_INTEGER frequency;
QueryPerformanceFrequency(&frequency);

int bufsize = 1 << 18;
long long statsize = 1ll << 34;
char* buf = malloc(bufsize);
uint64_t counter = 1;
int excess = 0;
while (counter < 9999999900000000)
{
LARGE_INTEGER start, stop;
QueryPerformanceCounter(&start);
for (int i = 0; i < statsize / bufsize; ++i)
{
memcpy(buf, buf + bufsize, excess);
counter = fill(buf + excess, bufsize - excess, counter, &excess);
e = WriteFile(f, buf, bufsize, &read, 0);
if (check_file)
FlushFileBuffers(f);
if (e == 0 || (int)read != bufsize)
{
e = GetLastError();
exit(1);
}
}
QueryPerformanceCounter(&stop);
printf("Throughput (GB/s): %f\n", statsize / (1 << 30) / time);
}

CloseHandle(f);
exit(0);
}

int main()
{
file_io();
}


This program will run faster than any other C or C++ entry, at least twice the speed. It is also noticeably faster than the last edit which was a bit buggy (blame vmplice). gcc as a compiler and SSSE3 (>= Core 2 Duo, Bulldozer/Bobcat) is a requirement.

There are two main optimizations that make this program run fast.

First, the digits are stored in a single XMM register with 128-bits, and since each character consumes 8-bits, one register can hold maximum 16 digits, which is more than enough. All operations to digits are done with one or several SIMD instructions, and the data always stays in a register before it gets copied to the buffer.

The PSHUFB instruction or _mm_shuffle_epi8 is the main source of speedup. It does the otherwise complicated byte reversal and shifting all bytes to one direction, in a single instruction. Both happen in the writeDigits function.

Second is vmsplice. I do not like the person who wrote this Linux-specific system call and probably the same person who wrote the documentation very badly. All would've been simple if such poorly-documented and unpredictable syscall didn't exist at all. I was really forced to use it because the speedup it could provide was too big. If you ever consider using this syscall, the following notes may help.

• The safest and probably intended use of vmplice is the sequence of mmap -> vmplice with SPLICE_F_GIFT -> munmap. You can call munmap directly after vmsplice retruns. This works in a similar way to asynchronous IO, and is very efficient in that sense. However since you are mmaping a new buffer every time, the buffer will unlikely be in cache.
• If you overwrite the buffer after a call to vmsplice, things can get very unpredictable. Whether SPLICE_F_GIFT is set or not doesn't make a difference, and I'm not even sure whether that flag does a thing at all. That vmsplice has returned does not mean that the pipe has consumed all of the buffer. That's why I said it is similar to asynchronous IO. It is not documented at all when it is safe to overwrite the buffer after vmsplice has returned. All I can say is it is somehow predictable when,
1. vmsplice is always called with the same buffer size.
2. The pipe size matches the buffer size.
3. The same thread writes to the buffer and calls vmsplice.
4. The buffer is not overwritten too fast... and I don't know how much is too fast, but for example if you overwrite the buffer right after vmsplice returns, things get unpredictable.

Compile the program with gcc -std=c11 -Ofast -mtune=native -mssse3 -fwhole-program -s.

#define _GNU_SOURCE
#include <stdbool.h>
#include <stdint.h>
#include <stdlib.h>
#include <string.h>
#include <fcntl.h>
#include <tmmintrin.h>

#define ALIGN 0x1000
#define SIZE (ALIGN << 8)
#define Z 0xf6f6f6f6f6f6f6f6u
static const union {uint64_t _[2]; __m128i m;} S[] = {
{{0x0706050403020100, 0x0f0e0d0c0b0a0908}},
{{0x0807060504030201, 0xff0f0e0d0c0b0a09}},
{{0x0908070605040302, 0xffff0f0e0d0c0b0a}},
{{0x0a09080706050403, 0xffffff0f0e0d0c0b}},
{{0x0b0a090807060504, 0xffffffff0f0e0d0c}},
{{0x0c0b0a0908070605, 0xffffffffff0f0e0d}},
{{0x0d0c0b0a09080706, 0xffffffffffff0f0e}},
{{0x0e0d0c0b0a090807, 0xffffffffffffff0f}},
{{0x0f0e0d0c0b0a0908, 0xffffffffffffffff}},
{{0xff0f0e0d0c0b0a09, 0xffffffffffffffff}},
{{0xffff0f0e0d0c0b0a, 0xffffffffffffffff}},
{{0xffffff0f0e0d0c0b, 0xffffffffffffffff}},
{{0xffffffff0f0e0d0c, 0xffffffffffffffff}},
{{0xffffffffff0f0e0d, 0xffffffffffffffff}},
{{0xffffffffffff0f0e, 0xffffffffffffffff}},
{{0xffffffffffffff0f, 0xffffffffffffffff}}
};

static __m128i inc(__m128i d) {
}

static __m128i inc_carry(__m128i d) {
}

static int writeDigits(char *b, __m128i d, int i) {
d = _mm_sub_epi64(d, _mm_set1_epi64x(0xc6c6c6c6c6c6c6c6));
d = _mm_shuffle_epi8(d, _mm_set_epi8(0, 1,  2,  3,  4,  5,  6,  7,
8, 9, 10, 11, 12, 13, 14, 15));
d = _mm_shuffle_epi8(d, S[i].m);
_mm_storeu_si128((__m128i *)b, d);
b[sizeof(d) - i] = '\n';
return sizeof(d) - i + 1;
}

static int writeFizz(char *b) {
uint64_t fizz = 0x0a7a7a6946;
memcpy(b, &fizz, sizeof(fizz));
return 5;
}

static int writeFizzBuzz(char *b) {
uint64_t fizzbuzz = 0x7a7a75427a7a6946;
memcpy(b, &fizzbuzz, sizeof(fizzbuzz));
b[8] = 0x0a;
return 9;
}

static int writeFizzAndBuzz(char *b) {
uint64_t fizznbuz = 0x7a75420a7a7a6946;
memcpy(b, &fizznbuz, sizeof(fizznbuz));
uint16_t zn = 0x0a7a;
memcpy(b + 8, &zn, sizeof(zn));
return 10;
}

static int writeBuzzAndFizz(char *b) {
uint64_t buzznfiz = 0x7a69460a7a7a7542;
memcpy(b, &buzznfiz, sizeof(buzznfiz));
uint16_t zn = 0x0a7a;
memcpy(b + 8, &zn, sizeof(zn));
return 10;
}

static __m128i fat(__m128i d) {
return _mm_or_si128(d, _mm_and_si128(_mm_set1_epi64x(Z),
_mm_cmpeq_epi8(d, _mm_setzero_si128())));
}

static void memcpy_fast(char *a, char *b, unsigned n) {
for (unsigned i = 0; i < n; i += sizeof(__m128i)) {
__m128i r = _mm_load_si128((__m128i *)(b + i));
_mm_store_si128((__m128i *)(a + i), r);
}
}

#define I d = inc(d)
#define IC d = inc_carry(d)
#define D I; p += writeDigits(p, d, i)
#define F I; p += writeFizz(p)
#define U I; p += writeFizzBuzz(p)
#define FNB I; I; p += writeFizzAndBuzz(p)
#define BNF I; I; p += writeBuzzAndFizz(p)
#define FIRST  D; D; F; D; BNF;  D; D; I; p += writeFizzAndBuzz(p)
#define SECOND D; F; D; D; U; D; D; F; D; p += writeBuzzAndFizz(p)
#define THIRD  I; D; D; FNB;  D; F; D; D; p += writeFizzBuzz(p)
#define CARRY(n) if (!((j + (n)) % 100000000)) IC; else I; d = fat(d)

int main() {
if (fcntl(1, F_SETPIPE_SZ, SIZE) != SIZE) abort();
_Alignas(ALIGN) char b[2][SIZE + ALIGN];
int f = 0;
char *p = b[f];
__m128i d = _mm_set1_epi64x(Z);
int i = sizeof(d) - 1;
long long j = 10, k = 10;
for (; j < 10000000000000000; j += 30) {
FIRST; CARRY(0);
if (j == k) {
k *= 10;
--i;
}
SECOND; CARRY(10);
THIRD; CARRY(20);
int n = p - b[f] - SIZE;
if (n >= 0) {
struct iovec v = {b[f], SIZE};
do {
int n = vmsplice(1, &v, 1, 0);
if (n < 0) abort();
v.iov_base = (char *)v.iov_base + n;
v.iov_len -= n;
} while (v.iov_len);
f = !f;
memcpy_fast(b[f], b[!f] + SIZE, n);
p = b[f] + n;
}
}
return 0;
}

• "I'm looking for a good way to test whether my program's output is correct" - See github.com/omertuc/fizzgolf/tree/main#verifying-submission Dec 23 '21 at 16:36
• @OmerTuchfeld Thanks for providing a testing module. I also made a pull request in github with the updated code. Dec 24 '21 at 3:51
• @OmerTuchfeld I couldn't run the test with the updated code. Now it requires to be piped to pv or cat like the asm entries. I tried adding taskset to the Makefile, but still coudn't run the test. The code runs fine on its own, but most certainly I'm doing something wrong using the test module. Dec 24 '21 at 4:19
• I believe the taskset commands are only there to improve performance, not to improve vmsplice compatibility. Not entirely sure why the test doesn't work for you because it works perfectly fine with the 2 asm entries which also use vmsplice. I'll have to look a bit into how vmsplice works then either fix your submission or fix the tester to make them compatible Dec 24 '21 at 12:07
• Okay seems like the tester not working on your newer version has nothing to do with vmsplice - the issue is that the tester is searching for the bytes \n1\n to detect the beginning of the fizzbuzz output (as opposed to junk compiler / make output) but your newer version starts printing numbers from 174871 rather than 1, so it never finds that string and ends up looping forever on it Dec 24 '21 at 14:56

My solution maintains a buffer with a batch of lines (6000 lines worked best on my system), and updates all the numbers in the buffer in a parallelisable loop. We use an auxiliary array nl[] to keep track of where each newline lies, so we have random access to all the numbers.

The addition is all in-place decimal character-by-character arithmetic, with no arithmetic division after the buffer is initialised (I could have created the buffer without division, too, but opted for shorter, readable code!). Every so often, when the number of digits rolls over, we have to stop and re-position all the numbers within the buffer (that's what the shuffle counter is for), and update the corresponding entries in nl[]; this happens more and more infrequently as we proceed.

I compiled using gcc -std=gnu17 -Wall -Wextra -fopenmp -O3 -march=native, and ran with OMP_NUM_THREADS=3 set in the environment (a different number of threads may be optimal on another host).

#include <stdatomic.h>
#include <stdio.h>              /* sprintf */
#include <string.h>             /* memset */
#include <unistd.h>

/* This is the single tunable you need to adjust for your platform */
#define chunk 6000 /* must be multiple of 3*5, with only one nonzero digit */
/* i.e. 3, 6 or 9 times an exact power of ten */

/* Select a number of digits to use.  If we produce one billion numbers
per second, then we'll finish all the 18-digit numbers in just 30
years.  24 digits should suffice until next geological epoch, at least. */
#define numlen 25               /* 24 decimal digits plus newline */

#define STR_(x) #x
#define STR(x) STR_(x)
#define chunk_str STR(chunk)

#define unlikely(e) __builtin_expect((e), 0)

char format[chunk * numlen];
char *nl[chunk+1];

int main()
{
/* Create the format string. */
/* We do this twice, as the numbers written first time round are
too short for the addition. */
for (int j = 0, n = 1;  j < 2;  ++j)
{
nl[0] = format;
char *p = format;
for (int i = 0;  i <= chunk;  ++i, ++n) {
if ((n % 15) == 0) {
p += sprintf(p, "FizzBuzz\n");
} else if ((n % 5) == 0) {
p += sprintf(p, "Buzz\n");
} else if ((n % 3) == 0) {
p += sprintf(p, "Fizz\n");
} else {
p += sprintf(p, "%d\n", n);
}
nl[i] = p;
}
write(1, format, nl[chunk] - format);
}

atomic_int shuffle = 0;
for (;;) {
#pragma omp parallel for schedule(static)
for (int i = 0;  i < chunk;  ++i) {
if (nl[i+1][-2] == 'z') {
/* fizz and/or buzz - not a number */
continue;
}
/* else add 'chunk' to the number */
static const int units_offset = sizeof chunk_str;
static const int digit = chunk_str[0] - '0';
char *p = nl[i+1] - units_offset;
*p += digit;
while (*p > '9') {
*p-- -= 10;
++*p;
}
if (unlikely(p < nl[i])) {
/* digit rollover */
++shuffle;
}
}
if (unlikely(shuffle)) {
char **nlp = nl + chunk;
char *p = *nlp;
char *dest = p + shuffle;
while (p < dest) {
if (*p == '\n') {
*nlp-- = dest + 1;
} else if (*p == '\n'+1) {
--*p;
*dest-- = '1';
*nlp-- = dest + 1;
}
*dest-- = *p--;
}
shuffle = 0;
}
write(1, format, nl[chunk] - format);
}
}

• Wouldn't memcpy(p,"Fizz\n",6); p+=5 be faster than sprintf? Jan 19 '21 at 22:05
• Optimal with OMP_NUM_THREADS=1 🤔. @EasyasPi tried that, didn't make a noticeable difference Jan 19 '21 at 22:22
• @EasyasPi, ITYM memcpy(p,"Fizz\n",5); to avoid pointlessly writing the null character each time? memcpy/sprintf tuning makes no measurable difference, as that's only the setup, outside the main loop. sprint() makes for more maintainable code. (I'm new to programming for all-out speed; in my day job that comes in third, after robustness and maintainability) Jan 20 '21 at 7:24
• @Omer: yes, the poor parallelisation was a disappointment. I tried some other parallelisations, too (separate array of formatted numbers, and putting the serialisation and writing into its own thread). If I'm to get any real benefit, I might have to go low-level and hand-craft the threads and their synchronisation (probably switch to C++ for that). Jan 20 '21 at 7:31
• I think I manage slightly better if I have atomic_int shuffle instead of the reduction. (time passes...) Yes, and I've updated to code that actually works faster in parallel, at last! Jan 20 '21 at 7:39

Trivial Rust

This one is just plain Rust without any tricks (no unsafe, no vmsplice(), no assembly), just a light loop unrolling. It manages to reach 6GiB/s on my laptop (XPS 15 i9) (it's wrong, see in comments), I'm curious to know how much it does on the reference hardware.

Compile with cargo build --release, run with ./target/release/fizzbuzz | pv >/dev/null

use std::error::Error;
use std::fmt::Write;
use std::io::Write as OtherWrite;

const LEN: usize = 1000000000;

fn main() -> Result<(), Box<dyn Error>> {
let stdout = std::io::stdout();
let mut stdout = std::io::BufWriter::new(stdout.lock());
let mut buffer = String::new();
buffer.reserve(80);
let mut n = 0usize;
while n < LEN {
write!(
&mut buffer,
r#"{}
{}
Fizz
{}
Buzz
Fizz
{}
{}
Fizz
Buzz
{}
Fizz
{}
{}
FizzBuzz
"#,
n + 1,
n + 2,
n + 4,
n + 7,
n + 8,
n + 11,
n + 13,
n + 14
)?;
stdout.write_all(buffer.as_bytes())?;
n += 15;
buffer.clear(); // forgot that ...
}
Ok(())
}
$$$$

• Getting 8.5GiB/s! Great throughput / complexity ratio. That puts its in third place I believe. This person promised 500$to whoever achieves this - news.ycombinator.com/item?id=29413879 Dec 5 '21 at 13:34 • Promised to donate to Watsi*, my bad Dec 5 '21 at 13:41 • Although I would suggest to first make sure that the output is correct and that the throughput holds up and doesn't decrease after a while in comparison to the C implementation before deciding on which is better, because the throughput is very close Dec 5 '21 at 13:45 • No sweat @OmerTuchfeld, I wasn't in there for the money anyway. Thanks a lot for the test, it would be nice to be on the podium for such a simple code. Dec 5 '21 at 13:56 • Unfortunately it's wrong. I forgot to clear the buffer, and now that I did it it kills the throughtput (I edited the code). Dec 5 '21 at 14:20 Python3 Interesting problem. I see most answers used static languages, except the Powershell answer, so far untimed. As I was curious about how dynamic languages would fare in this task, I wrote a simple implementation in Python. I ran it under GNU/Linux Mint 20.02 64-bit, using default Python3 (3.8.10) and pypy (7.3.1) in the repositories. Processor model in my machine: AMD Athlon(tm) X4 750 Quad Core. Initial version The initial version just keeps a carousel running in sync to a counter, where the carousel carries either False or the results different from the current count, in the proper positions. The derailleur function selects what is to be printed. from itertools import cycle, count def derailleur(counter, carousel): if not carousel: return counter return carousel def main(): carousel = cycle([0,0,'Fizz',0,'Buzz','Fizz',0,0,'Fizz','Buzz',0,'Fizz',0,0,'FizzBuzz']) counter = count(1) f = map(print, map(derailleur, counter, carousel)) while 1: next(f) main()  In my machine it ran at about 14,2MiB/s under CPython, and got a modest boost up to 25,0MiB/s under pypy: user@Desktop:~$ python3 fizzbuzz.py | pv > /dev/null
^C21MiB 0:00:30 [14,2MiB/s] [                            <=>                   ]
Traceback (most recent call last):
File "fizzbuzz.py", line 15, in <module>
main()
File "fizzbuzz.py", line 13, in main
next(f)
File "fizzbuzz.py", line 3, in derailleur
def derailleur(counter, carousel):
KeyboardInterrupt
Exception ignored in: <_io.TextIOWrapper name='<stdout>' mode='w' encoding='utf-8'>
BrokenPipeError: [Errno 32] Broken pipe

user@Desktop:~$pypy3 fizzbuzz.py | pv > /dev/null ^C57MiB 0:00:30 [25,0MiB/s] [ <=> ] Traceback (most recent call last): File "fizzbuzz.py", line 15, in <module> main() File "fizzbuzz.py", line 13, in main next(f) KeyboardInterrupt  Chunking version Afterwards I modified the initial version to print the results in chunks, instead of one by one. from itertools import cycle, count def derailleur(counter, carousel): if not carousel: return counter return carousel def main(): carousel = cycle([0,0,'Fizz',0,'Buzz','Fizz',0,0,'Fizz','Buzz',0,'Fizz',0,0,'FizzBuzz']) counter = map(str, count(1)) f = map(derailleur, counter, carousel) while 1: print('\n'.join([next(f) for _ in range(256)])) main()  Now under CPython it runs a bit faster, at about 19,6MiB/s. But under pypy it receives a large boost, achieving 84,9MiB/s, within half the speed of the naive implementation written in C reported by the OP (170MiB/s): user@Desktop:~$ python3 chunking_fizzbuzz.py | pv > /dev/null
^C84MiB 0:00:30 [19,6MiB/s] [                            <=>                   ]
Traceback (most recent call last):
File "chunking_fizzbuzz.py", line 15, in <module>
main()
File "chunking_fizzbuzz.py", line 13, in main
print('\n'.join([next(f) for _ in range(256)]))
KeyboardInterrupt

user@Desktop:~$pypy3 chunking_fizzbuzz.py | pv > /dev/null ^C49GiB 0:00:30 [84,9MiB/s] [ <=> ] Traceback (most recent call last): File "chunking_fizzbuzz.py", line 15, in <module> main() File "chunking_fizzbuzz.py", line 13, in main print('\n'.join([next(f) for _ in range(256)])) File "chunking_fizzbuzz.py", line 13, in <listcomp> print('\n'.join([next(f) for _ in range(256)])) KeyboardInterrupt  I'm curious to see how much this result can be improved upon, for Python, and how other dynamic languages perform. • Welcome to Code Golf and nice first answer! Nov 11 '21 at 20:41 Julia 1.7 @inbounds function inc!(l::Vector{UInt8}, n, i = length(l)-1) if i > 0 l[i] += n if l[i] > 0x39 # '9' l[i] -= 0x0A # 10 inc!(l, 0x01, i-1) end else pushfirst!(l, 0x31) # '1' end return nothing end function main(maxi = typemax(Int), N = 2^16) io = IOBuffer(; sizehint=N) a = UInt8['1', '\n'] sizehint!(a, 100) for j in 1:N:maxi s = take!(io) write(stdout, s) l = ll = 0 while ll+l+8 < N l = 0 l += write(io, a) inc!(a, 0x01) l += write(io, a) l += write(io, "Fizz\n") inc!(a, 0x02) l += write(io, a) l += write(io, "Buzz\nFizz\n") inc!(a, 0x03) l += write(io, a) inc!(a, 0x01) l += write(io, a) l += write(io, "Fizz\nBuzz\n") inc!(a, 0x03) l += write(io, a) l += write(io, "Fizz\n") inc!(a, 0x02) l += write(io, a) inc!(a, 0x01) l += write(io, a) l += write(io, "FizzBuzz\n") inc!(a, 0x02) ll += l end end end main()  The buffer of size N is optimized for my machine. I get about 350-400 MiB/s in WSL, it might be better on a real Linux machine (let's not talk about the performance in windows). I got slightly better results with julia 1.7 than with 1.6. Try it online! Updated code: This code (no more strings) gets me to 4.8GiB/s. Can't break the 5 GiB/s barrier on the M1 with the naive implementation :-(. Also, larger segments, even more memory. However, this has better throughput than just writing empty arrays to the stdout (even with draining to the direct byte buffer). import java.io.FileDescriptor; import java.io.FileOutputStream; import java.nio.ByteBuffer; import java.nio.charset.StandardCharsets; import java.util.LinkedList; import java.util.Optional; import java.util.Queue; import java.util.concurrent.CompletableFuture; import java.util.concurrent.ForkJoinPool; import java.util.function.Consumer; import java.util.function.Supplier; import java.util.stream.IntStream; public class FizzBuzz { static final String NEWLINE = String.format("%n"); static byte[] fizzbuzz = "fizzbuzz\n".getBytes(StandardCharsets.ISO_8859_1); static byte[] fizz = "fizz\n".getBytes(StandardCharsets.ISO_8859_1); static byte[] buzz = "buzz\n".getBytes(StandardCharsets.ISO_8859_1); static byte newline = "\n".getBytes(StandardCharsets.ISO_8859_1)[0]; static byte zero = (byte) '0'; static int fizzBuzz(long src, byte[] destination, int destinationPos, byte[] digitsHelper, int digitsHelperLast) { if (src % 15 == 0) { System.arraycopy(fizzbuzz, 0, destination, destinationPos, fizzbuzz.length); return destinationPos + fizzbuzz.length; } else if (src % 5 == 0) { System.arraycopy(fizz, 0, destination, destinationPos, fizz.length); return destinationPos + fizz.length; } else if (src % 3 == 0) { System.arraycopy(buzz, 0, destination, destinationPos, buzz.length); return destinationPos + buzz.length; } else { do { digitsHelper[digitsHelperLast--] = (byte) (zero + src % 10); src /= 10; } while (src != 0); int len = digitsHelper.length - digitsHelperLast; System.arraycopy(digitsHelper, digitsHelperLast, destination, destinationPos, len); return destinationPos + len; } } record Pair(byte[] item1, ByteBuffer item2) { } public static void main(String[] argv) throws Exception { final var channel = new FileOutputStream(FileDescriptor.out).getChannel(); final Consumer<ByteBuffer> writeToChannel = bb -> { try { while (bb.hasRemaining()) { channel.write(bb); } } catch (Exception ex) { ex.printStackTrace(); System.exit(1); } }; final var segment = 10_000_000; final var arralloc = 20 * segment; final Queue<CompletableFuture<Pair>> queue = new LinkedList<> (IntStream.range(0, Optional.of(ForkJoinPool.getCommonPoolParallelism() - 1).filter(i -> i > 0).orElse(1)) .mapToObj(i -> CompletableFuture.completedFuture(new Pair(new byte[arralloc], ByteBuffer.allocateDirect(arralloc)))).toList()); CompletableFuture<?> last = CompletableFuture.completedFuture(null); final Supplier<Pair> supplier = () -> { try { return queue.poll().get(); } catch (Exception ex) { ex.printStackTrace(); System.exit(1); return null; } }; var nr = 0L; while (true) { final var start = nr; final var end = nr + segment; final var finalLast = last; var cf = CompletableFuture.completedFuture(supplier.get()) .thenApplyAsync(p -> { var arr = p.item1(); var bb = p.item2(); var pos = 0; var digitsHelper = new byte[20]; digitsHelper[19] = newline; for (long l = start; l < end; l++) { pos = fizzBuzz(l, arr, pos, digitsHelper, 18); } bb.clear(); bb.put(arr, 0, pos); bb.flip(); return p; }) .thenCombineAsync(finalLast, (p, v) -> { try { channel.write(p.item2()); } catch (Exception ex) { ex.printStackTrace(); System.exit(1); } return p; }); queue.add(cf); last = cf; nr = end; } } }  Regular JAVA, no tricks on fizzbuzz, needs quite a bit of memory and benefits from multiple threads. On my M1 Pro w/ JAVA 17 I am at 4.3GiB/s (9 threads + 1 main). ByteBuffer writes max out the channel at ~ 5GiB/s piped via pv (without fizzbuzz). import java.io.FileDescriptor; import java.io.FileOutputStream; import java.nio.ByteBuffer; import java.nio.charset.StandardCharsets; import java.util.LinkedList; import java.util.Optional; import java.util.Queue; import java.util.concurrent.CompletableFuture; import java.util.concurrent.ForkJoinPool; import java.util.function.Consumer; import java.util.function.Supplier; import java.util.stream.IntStream; public class FizzBuzz { static final String NEWLINE = String.format("%n"); static StringBuilder fizzBuzz(long src, StringBuilder destination) { if (src % 15 == 0) { destination.append("fizzbuzz"); } else if (src % 5 == 0) { destination.append("fizz"); } else if (src % 3 == 0) { destination.append("buzz"); } else { destination.append(src); } return destination.append(NEWLINE); } record Pair<T1, T2>(T1 item1, T2 item2) { } public static void main(String[] argv) throws Exception { final var channel = new FileOutputStream(FileDescriptor.out).getChannel(); final Consumer<ByteBuffer> writeToChannel = bb -> { try { while (bb.hasRemaining()) { channel.write(bb); } } catch (Exception ex) { ex.printStackTrace(); System.exit(1); } }; final Queue<CompletableFuture<Pair<StringBuilder, ByteBuffer>>> queue = new LinkedList<> (IntStream.range(0, Optional.of(ForkJoinPool.getCommonPoolParallelism() - 1).filter(i -> i > 0).orElse(1)) .mapToObj(i -> CompletableFuture.completedFuture(new Pair<>(new StringBuilder(20 * 1024 * 1024), ByteBuffer.allocateDirect(41 * 1024 * 1024)))).toList()); CompletableFuture<?> last = CompletableFuture.completedFuture(null); final Supplier<Pair<StringBuilder, ByteBuffer>> supplier = () -> { try { return queue.poll().get(); } catch (Exception ex) { ex.printStackTrace(); System.exit(1); return null; } }; var nr = 0L; var segment = 1_000_000L; while (true) { final var start = nr; final var end = nr + segment; final var finalLast = last; var cf = CompletableFuture.completedFuture(supplier.get()) .thenApplyAsync(p -> { var sb = p.item1(); var bb = p.item2(); sb.delete(0, sb.length()); for (long l = start; l < end; l++) { fizzBuzz(l, sb); } bb.clear(); bb.put(sb.toString().getBytes(StandardCharsets.UTF_8)); return p; }) .thenCombineAsync(finalLast, (p, v) -> { try { channel.write(p.item2().flip()); } catch (Exception ex) { ex.printStackTrace(); System.exit(1); } return p; }); queue.add(cf); last = cf; nr = end; } } }  • Doesn't even seem to reach 3GiB/s on my machine. I ran it with /usr/lib/jvm/java-17-openjdk-17.0.1.0.12-1.rolling.fc35.x86_64/bin/java FizzBuzz.java | pv > /dev/null - let me know if there's some other way I should be running this Dec 8 '21 at 23:22 • Probably not. I get the same values (4.3) with both oracle jdk and openjdk so I don't think that that will make a difference for you, either. The thread count is set by Optional.of(ForkJoinPool.getCommonPoolParallelism() - 1).filter(i -> i > 0).orElse(1) - and that can be manually replaced with the number of cores in case it gets it wrong (cores - 1). Dec 8 '21 at 23:44 • No matter what value I give of(...) I can't seem to go past 2.6GiB/s. Must be some M1 pro magic giving you this throughput Dec 9 '21 at 0:05 • It probably is the M1. What would you get on pv for this: import java.io.FileDescriptor; import java.io.FileOutputStream; import java.nio.ByteBuffer; public class Test { public static void main(String[] argv) throws Exception { var channel = new FileOutputStream(FileDescriptor.out).getChannel(); var buffer = ByteBuffer.allocateDirect(41 * 1024 * 1024); while(true) { while(buffer.hasRemaining()) { channel.write(buffer); } buffer.clear(); } }} ? Just curios what maxes first. Dec 9 '21 at 0:23 • Welcome to Code Golf, and nice answer! Dec 9 '21 at 4:27 Java 17 Compile and run with java FizzBuzzAsync.java. It requires a warmup period, but I don't know why. Throughput on my 6-years-old laptop with 8 processors: 3.6 GiB/s (actually, between 210 and 220 GiB in 1 minute). The algorithm is correct for numbers up to 2^63-1. import java.io.FileDescriptor; import java.io.FileOutputStream; import java.nio.ByteBuffer; import java.util.Arrays; import java.util.concurrent.Callable; import java.util.concurrent.Executors; import java.util.concurrent.Future; import java.util.concurrent.LinkedBlockingQueue; import java.util.stream.IntStream; import static java.nio.charset.StandardCharsets.US_ASCII; public final class FizzBuzzAsync { private static final long STARTING_NUMBER = 10L; private static final long INCREMENT = 5_400L; private static final long SIZE_FOR_30 = 2 * (17 * 8 + 47); private static final int[] BASE_IDX = {-1, 6, 8, 19, 21, 28, 40, 42, 54, 61, 63, 74, 76, 83, 95, 97}; public static void main(String[] args) { new FizzBuzzAsync().run(); } private void run() { var threads = Runtime.getRuntime().availableProcessors(); var queue = new LinkedBlockingQueue<Future<ByteBuffer>>(threads); var executor = Executors.newFixedThreadPool(threads); try (var channel = new FileOutputStream(FileDescriptor.out).getChannel()) { var counter = STARTING_NUMBER; for (var i = 0; i < threads; i++) { var buffer = ByteBuffer.allocateDirect((int) (INCREMENT * SIZE_FOR_30 / 30L)); queue.offer(executor.submit(new Task(counter, buffer))); counter += INCREMENT; } channel.write(ByteBuffer.wrap("1\n2\nFizz\n4\nBuzz\nFizz\n7\n8\nBuzz\nFizz\n".getBytes(US_ASCII))); while (true) { var buffer = queue.poll().get(); buffer.flip(); channel.write(buffer); buffer.clear(); queue.offer(executor.submit(new Task(counter, buffer))); counter += INCREMENT; } } catch (Exception e) { e.printStackTrace(System.err); } finally { executor.shutdown(); } } record Task(long startingNumber, ByteBuffer buffer) implements Callable<ByteBuffer> { @Override public ByteBuffer call() { var n = startingNumber; var idx = Arrays.copyOf(BASE_IDX, BASE_IDX.length); var nextPowerOf10 = 10L; for (; nextPowerOf10 < n; nextPowerOf10 *= 10L) { for (var i = 0; i < idx.length; ) { idx[i] += ++i; } } byte[] template = null; for (var limit = n + INCREMENT; n < limit; n += 30) { if (template == null | n == nextPowerOf10) { if (n == nextPowerOf10) { nextPowerOf10 *= 10; for (var i = 0; i < idx.length; ) { idx[i] += ++i; } } template = ((n + 1) + "\nFizz\n" + (n + 3) + "\n" + (n + 4) + "\nFizzBuzz\n" + (n + 6) + "\n" + (n + 7) + "\nFizz\n" + (n + 9) + "\nBuzz\nFizz\n" + (n + 12) + "\n" + (n + 13) + "\nFizz\nBuzz\n" + (n + 16) + "\nFizz\n" + (n + 18) + "\n" + (n + 19) + "\nFizzBuzz\n" + (n + 21) + "\n" + (n + 22) + "\nFizz\n" + (n + 24) + "\nBuzz\nFizz\n" + (n + 27) + "\n" + (n + 28) + "\nFizz\nBuzz\n").getBytes(US_ASCII); } else { for (var i = 0; i < idx.length; ) { var pos = idx[i++]; template[pos] += 3; while (template[pos] > '9') { template[pos] -= 10; template[--pos]++; } } } buffer.put(template); } return buffer; } } }  • TODO: learn the vector API to increase speed. Dec 3 '21 at 15:00 Just tried the following Kotlin, compiled to Kotlin Native: fun main(){ for (i in 1..1_000_000_000){ when { (i % 3 == 0 && i % 5 == 0) -> println("FizzBuzz") i % 3 == 0 -> println("Fizz") i % 5 == 0 -> println("Buzz") else -> println("$i")
}
}
}


2.91MiB/s

Machine specs:

• Linux 5.14.18
• i7-8550U
• 16GB Ram
• Welcome to Code Golf, and nice answer! Dec 2 '21 at 14:27

Python3

I ported Neil+Kamila answer to python3

from os import write

buf = bytearray(256)
out = bytearray(65536 + 4096)

init = b"1\n2\nFizz\n4\nBuzz\nFizz\n7\n8\nFizz\n"

out[:len(init)] = init

fmt = "Buzz\n{}1\nFizz\n{}3\n{}4\nFizzBuzz\n{}6\n{}7\nFizz\n{}9\nBuzz\nFizz\n{}2\n{}3\nFizz\nBuzz\n{}6\nFizz\n{}8\n{}9\nFizzBuzz\n{}1\n{}2\nFizz\n{}4\nBuzz\nFizz\n{}7\n{}8\nFizz\n"

t = 30
i = 1
j = 1

for l in range(1, 20):
txt = fmt.format(i, i, i, i, i, i, i + 1, i + 1, i + 1, i + 1, i + 1, i + 2, i + 2, i + 2, i + 2, i + 2).encode()
buf[:len(txt)] = txt

i *= 10
while j < i:
out[t:t+len(txt)] = buf[:len(txt)]
t += len(txt)
if t >= 65536:
u = write(1, out[:65536])
while u < 65536:
u += write(1, out[u:65536])

t -= 65536
out[:t] = out[65536:65536+t]

q = 0
for z in (4, 7, 2, 11, 2, 7, 12, 2, 12, 7, 2, 11, 2, 7, 12, 2):
q += z + l
p = q
if buf[p] < 55:
buf[p] += 3
else:
buf[p] -= 7
p -= 1
while buf[p] == 57:
buf[p] = 48
p -= 1

buf[p] += 1

j += 3


On my laptop with i5-8300H $python3 fizzbuzz.py | pv > /dev/null results in throughput of 48 MiB/s on average. While running the same code with pypy3 gives me up to 820 MiB/s. • Welcome to Code Golf, and nice first answer! Dec 2 '21 at 20:35 I wrote two versions in C# 10: one using "vanilla" language features and one based on Isaac's implementation. In the optimized version, I exploited the fact that an exact length memcpy is not necessary if the buffer is large enough. It's limited to 10^16 iterations, and the first 10 results have a leading 0 (I just thought it would be worth posting it anyway). Results: OS: Ubuntu 20.04.3 LTS CPU: Intel(R) Xeon(R) Platinum 8171M CPU @ 2.60GHz (note: this is a GitHub Actions worker) Author Lang Avg. Speed ----------------------------------------------- Paolo Bonzini C 11.38 GiB/s Isaac G. C 1.90 GiB/s Neil C 1.83 GiB/s Kamila Szewczyk C 1.44 GiB/s Daniel C# (opt) 1006.00 MiB/s Olivier Grégoire Java 242.62 MiB/s Daniel C# (simpl) 125.125 MiB/s  Simple version using System.Text; var sb = new StringBuilder(); ulong c = 0; while (true) { while (sb.Length < 1024 * 1024 * 4) { sb.Append($"{c + 1}\n{c + 2}\nFizz\n{c + 4}\nBuzz\nFizz\n{c + 7}\n{c + 8}\nFizz\nBuzz\n{c + 11}\nFizz\n{c + 13}\n{c + 14}\nFizzBuzz\n");
c += 15;
}
Console.Out.Write(sb);
sb.Clear();
}


"Optimized" version

using System.Runtime.CompilerServices;
using System.Runtime.Intrinsics;

using var stdout = Console.OpenStandardOutput();

var buf = new byte[1024 * 1024 * 4];
var counter = new Counter();

while (true) {
int pos = 0;
for (int i = 0; i < 4096; i++) {
pos = Fill30(buf, pos, ref counter);
}
stdout.Write(buf.AsSpan(0, pos));
}

int Fill30(byte[] buf, int pos, ref Counter ctr)
{
var Fizz_ = 0x0A_7A_7A_69_46ul;
var Fizz_Buzz_ = Vector128.Create(0x7A_75_42_0A_7A_7A_69_46ul, 0x0A_7A).AsByte();
var Buzz_Fizz_ = Vector128.Create(0x7A_69_46_0A_7A_7A_75_42ul, 0x0A_7A).AsByte();
var FizzBuzz__ = Vector128.Create(0x7A_7A_75_42_7A_7A_69_46, 0x0Aul).AsByte();

//0..9
var prefix = ctr.Get();
int prefixLen = ctr.NumDigits;
ctr.Inc();

//10..19
prefix = ctr.Get();
prefixLen = ctr.NumDigits;
ctr.Inc();

//20..29
prefix = ctr.Get();
prefixLen = ctr.NumDigits;
ctr.Inc();

return pos;

{
//ref byte ptr = ref Unsafe.Add(ref MemoryMarshal.GetArrayDataReference(buf), pos);
//Unsafe.WriteUnaligned(ref ptr, val);
Unsafe.WriteUnaligned(ref buf[pos], val);
pos += len;
}
{
Add((short)(0x0A_30 + digit), 2); // "<digit>\n" ascii in little endian
}
}

unsafe struct Counter
{
public const int MAX_DIGITS = 16;
public fixed byte Digits[MAX_DIGITS * 2];
public int NumDigits;

public Counter()
{
Unsafe.InitBlock(ref Digits[0], (byte)'0', MAX_DIGITS);
NumDigits = 1;
}
public void Inc()
{
int i = MAX_DIGITS - 1;
for (; i >= 0; i--) {
if (Digits[i] != (byte)'9') {
Digits[i]++;
break;
}
Digits[i] = (byte)'0';
}
NumDigits = Math.Max(NumDigits, MAX_DIGITS - i);
}
public Vector128<byte> Get()
{
ref byte ptr = ref Digits[MAX_DIGITS - NumDigits];
}
}
$$$$


• Use f-strings not .format() (because that'd add a function call) in the derailleur() function

• Use os.write() instead of print(), because print() is slow (not that os.write is much faster, but it is faster)

It ends up looking like this:

from itertools import cycle, count
from os import write

def derailleur(counter, carousel):
return f"{carousel if carousel else counter}\n"

def main():
carousel = cycle([0, 0, "Fizz", 0, "Buzz", "Fizz", 0, 0, "Fizz", "Buzz", 0, "Fizz", 0, 0, "FizzBuzz"])
counter = count(1)
f = map(derailleur, counter, carousel)
while True:
write(1,  "".join( [ next(f) for _ in range(8192) ] ).encode("utf-8") )

main()


I also played a bit with extending the range, but didn't get a significant speedup from that.

Under WSL2 on a 7200u w/ 16GiB of RAM I'm getting about a 10% to 20% better throughput, but the readings are super noisy (it starts at 15 MiB/s, jumps to 25, back to 20, etc...).

I don't have a Linux machine at hand to test this under a better environment, but I don't expect these changes to make a significant impact. After a few measurements, either using print() or os.write(), about 95% of the time is spent writing the output, and I haven't done much to address that with my changes.

Edit: fixed a few typos and tested on PyPy3 under the same environment:

• Nearing 100 MiB/s using os.write()
• Slightly faster 108 MiB/s using print()

Edit 2: moar PyPy!

If we throw CPython performance out of the window, we can go with this pretty straightforward code:

def fizzbuzz(x: int) -> str:
if x % 15 == 0: return "FizzBuzz"
if x % 5  == 0: return "Buzz"
if x % 3  == 0: return "Fizz"
return f"{x}"

def main():
c = 0
while True:
print( "\n".join( fizzbuzz( c + i) for i in range(8192) ) )
c += 8192

main()


And that is netting about 148 MiB/s on my PC, using PyPy. So about 6x faster than ksousas CPython version. Running it under python3 gives me 22-24 MiB/s, which is close to where I started. Again, I'm fairly certain on my environment writing to stdout dominates and there's very little to be done unless that bottleneck can be addressed.

Java: On my 11th gen intel laptop

• The OP's simple C version: 25 MiB/s
• My simple Java version: 140 MiB/s
• python neil+kamila: 8 MiB/s
• python ksoua chunking: 4.33 MiB/s
• Jan's C version: 1.0 GiB/s

Sometimes simple is better!

To the guy that did a threaded C++ version: wow!

public class FizzBuzz {

public static void main(String[] args) {

long maxBufLen=4096<<3;

var sb=new StringBuilder((int)maxBufLen + 10);
for (long i = 1; ; i+=1) {

if ((i % 3 == 0) && (i % 5 == 0)) {
sb.append("FizzBuzz\n");
} else if (i % 3 == 0) {
sb.append("Fizz\n");
} else if (i % 5 == 0) {
sb.append("Buzz\n");
} else {
sb.append(i).append("\n");
}

if(sb.length()>maxBufLen) {
System.out.print(sb);
sb.setLength(0);
}
}

}

}


Copied Ioan's threaded version (not the way I'd do it but it works...) and added some efficiencies:

750MiB/s same laptop


import java.io.FileDescriptor;
import java.io.FileOutputStream;
import java.nio.ByteBuffer;
import java.nio.CharBuffer;
import java.nio.channels.FileChannel;
import java.nio.charset.CharsetEncoder;
import java.nio.charset.StandardCharsets;
import java.util.Queue;
import java.util.concurrent.CompletableFuture;
import java.util.concurrent.ForkJoinPool;
import java.util.function.Supplier;
import java.util.stream.IntStream;

public class FizzBuzzIoan {

static final String NEWLINE = String.format("%n");
static final CharsetEncoder encoder = StandardCharsets.US_ASCII.newEncoder();

static void fizzBuzz(long src, StringBuilder destination) {

if (src % 15 == 0) {
destination.append("FizzBuzz"+NEWLINE);
} else if (src % 5 == 0) {
destination.append("Fizz"+NEWLINE);
} else if (src % 3 == 0) {
destination.append("Buzz"+NEWLINE);
} else {
destination.append(src).append(NEWLINE);
}

}

record Pair<T1, T2>(T1 item1, T2 item2) {
}

public static void main(String[] args) throws Exception {

try (var fis= new FileOutputStream(FileDescriptor.out)) {
var channel = fis.getChannel();
doIt(channel);
}
}

static void doIt(FileChannel channel) {

var max=Math.max(ForkJoinPool.getCommonPoolParallelism(), 1);

final Queue<CompletableFuture<Pair<StringBuilder, ByteBuffer>>> queue =
IntStream.range(0, max).mapToObj(i ->
CompletableFuture.completedFuture(
new Pair<>(new StringBuilder(20 * 1024 * 1024), ByteBuffer.allocateDirect(41 * 1024 * 1024)))).toList()
);

CompletableFuture<?> last = CompletableFuture.completedFuture(null);

final Supplier<Pair<StringBuilder, ByteBuffer>> supplier = () -> {
try {
return queue.poll().get();
} catch (Exception ex) {
ex.printStackTrace();
System.exit(1);
return null;
}
};

var nr = 0L;
var segment = 1_000_000L;

while (true) {
final var start = nr;
final var end = nr + segment;

final var finalLast = last;

var cf = CompletableFuture.completedFuture(supplier.get())
.thenApplyAsync(p -> {
var sb = p.item1();
var bb = p.item2();

sb.setLength(0);

for (long l = start; l < end; l++) {
fizzBuzz(l, sb);
}
bb.clear();

try {
var cb=CharBuffer.wrap(sb);
encoder.encode(cb, bb, true);

} catch(Exception ex) {
throw new RuntimeException(ex);
}
return p;

})
.thenCombineAsync(finalLast, (p, v) -> {
try {
channel.write(p.item2().flip());
} catch (Exception ex) {
ex.printStackTrace();
System.exit(1);
}
return p;
});

last = cf;

nr = end;

}

}

}

$$$$

• Welcome to Code Golf, and nice first answer! Dec 3 '21 at 0:07

A C++ program for Linux. I use the same method of arithmetic as in my C answer, but here I create a team of threads by hand rather than using OpenMP.

We divide the problem into ranges of numbers, so that we don't have to touch the low-order digits each iteration.

The workers are arranged in a circular chain, and each is responsible for a subrange. We do all our addition while other threads are writing, then take a turn at writing. In order to write an exact number of pages with each write, there's generally one write that straddles two workers, so we combine writes using writev(), temporarily blocking the other worker from beginning its arithmetic. The timing diagram looks like this:

    +--------------------------------------------+
|                                            |
|   wait                                     |
------->                                         |
|      write (end of prev, start of this)    |          +--------------------------------------------+
<-------                                         |          |                                            |
|      write (just this one)                 |          |   wait                                     |
|                                      -------------------->                                         |
|                                  wait      |          |      write (end of prev, start of this)    |
|                                      <--------------------                                         |
|      update each number                    |          |      write (just this one)                 |
|                                            |          |                                      ------+----->
+--------------------------------------------+          |                                  wait      |
|                                      <------------
|      update each number                    |
|                                            |
+--------------------------------------------+


On my workstation (8 thread Ivybridge), I measured at around 90% the throughput of cat /dev/zero, or about ¾ that of dd if=/dev/zero bs=64K.

#include <cassert>
#include <condition_variable>
#include <iostream>
#include <memory>
#include <mutex>
#include <vector>

#include <unistd.h>
#include <sys/sysinfo.h>
#include <sys/uio.h>

// Select a number of decimal digits to use.  If we produce one billion
// numbers per second, then we'll finish all the 18-digit numbers in
// just 30 years.  24 digits should suffice until the next geological
// epoch, at least.
static constexpr int numlen = 25; // 24 digits plus newline

static constexpr int write_size = 0x10000; // this is fastest on my system

static_assert('0' - '\n' > 1, "Character coding incompatible with the arithmetic");

#define unlikely(e) __builtin_expect((e), 0)

struct worker
{
// Storage for the character string we maintain
std::string lines{""};
// Iterators to each newline in 'lines'
std::vector<std::string::iterator> nl{};

int units_offset{};         // significant figures in the step
char digit{};               // the single non-zero digit of step

// We coordinate with the next and previous threads in the ring.
// The iovec is used for writing a block of lines that straddles
// the previous thread and this one.
std::mutex mutex{};
std::condition_variable cv{};
struct iovec iov[2] = { { 0, 0 }, { 0, 0 } };
worker *next{};

// The functions
worker(std::size_t first, std::size_t count, std::size_t step);
worker(const worker&) = delete;
worker& operator=(const worker&) = delete;
void loop();
};

static constexpr auto buf_len(std::size_t digits, std::size_t count)
{
// each group of 15 lines has 8 numbers and 39 chars of Fizz and Buzz
return (8 * digits + 39) * count / 15;
}

{
// We need each thread to produce at least write_size each iteration
for (std::size_t tens = 1000;  tens < 10'000'000;  tens *= 10) {
auto const digits = std::snprintf(0, 0, "%zu", tens);
for (std::size_t digit = 3;  digit < 10;  digit += 3) {
if (buf_len(digits, digit * tens) / thread_count > write_size) {
return digit * tens;
}
}
}
return 9'000'000; // fallback (perhaps we should limit thread count?)
}

int main()
{
// How many threads will we have?
auto const nprocs = get_nprocs();
auto const step = optimal_step(nprocs);

// Write output the slow way, until we have enough digits for the
// format strings.
auto n = 1;
const auto step_len = std::to_string(step).size();
// Finish the loop just before a FizzBuzz, so that the lines buffer
// doesn't start with a number (we need a newline preceding for it
// to overflow correctly).
for (;  std::to_string(n).size() <= step_len;  ++n) {
if ((n % 15) == 0) {
std::cout << "FizzBuzz\n";
} else if ((n % 5) == 0) {
std::cout << "Buzz\n";
} else if ((n % 3) == 0) {
std::cout << "Fizz\n";
} else {
std::cout << n << '\n';
}
}
std::cout.flush();          // use Unix write() from here on

// create the workers
auto workers = std::vector<std::unique_ptr<worker>>{};
workers.reserve(nprocs);
for (int i = 0;  i < nprocs;  ++i) {
auto const a = i * step / nprocs;
auto const z = (i+1) * step / nprocs;
workers.emplace_back(std::make_unique<worker>(n + a, z - a, step));
}
// and connect them in a loop
workers.back()->next = workers.front().get();
for (int i = 1;  i < nprocs;  ++i) {
workers[i-1]->next = workers[i].get();
}

// a thread for each worker
auto const loop = [](worker *w) { w->loop(); };
for (auto const& w: workers) {
}

// start them writing
auto &first = workers.front();
{
std::unique_lock lock{first->mutex};
first->iov[0] = { &first->lines.front(), 0 };
}
first->cv.notify_one();

}

worker::worker(std::size_t first, std::size_t count, std::size_t step)
{
auto s = std::to_string(step);
units_offset = s.size() + 1;
digit = s.front() - '0';

lines.reserve(buf_len(numlen, count) + 1);
nl.reserve(count);
assert(lines.capacity() >= write_size);

for (auto n = first;  n < first + count;  ++n) {
if ((n % 15) == 0) {
lines += "FizzBuzz\n";
} else if ((n % 5) == 0) {
lines += "Buzz\n";
} else if ((n % 3) == 0) {
lines += "Fizz\n";
} else {
lines += std::to_string(n) + '\n';
}
nl.push_back(lines.end());
}
assert(lines.size() >= write_size);

// Second half of a straddle write is always from the beginning of our buffer.
iov[1].iov_base = &lines.front();
}

void worker::loop()
{
assert(next);
for (;;) {
{
// We can write when the previous thread passes its buffer offcut.
std::unique_lock lock{mutex};
cv.wait(lock, [this]{ return iov[0].iov_base; });

iov[1].iov_len = write_size - iov[0].iov_len;
assert(iov[1].iov_len < lines.size());
writev(1, iov, 2);
// Tell the previous thread that we've finished with its buffer.
iov[0].iov_base = 0;
}
cv.notify_one();

auto p = lines.begin() + iov[1].iov_len;
while (unlikely(p + write_size < lines.end())) {
::write(1, &*p, write_size);
p += write_size;
}

{
// Tell the next thread that we have leftover data for it to write.
std::unique_lock lock{next->mutex};
next->iov[0] = { &*p, static_cast<std::size_t>(lines.end() - p) };
}
next->cv.notify_one();

{
// Now wait until next worker has written, and released buffer back to us.
std::unique_lock lock{next->mutex};
next->cv.wait(lock, [this]{ return !next->iov[0].iov_base; });
}

// Update the numbers in the buffer.
auto rollover = 0u;
for (auto const& i: nl) {
if (i[-2] == 'z') {
// fizz and/or buzz - not a number
continue;
}
// else add 'step' to the number
auto p = i - units_offset;
*p += digit;
while (*p > '9') {
*p-- -= 10;
++*p;
}
if (unlikely(*p == '\n'+1)) {
// digit rollover
++rollover;
}
}
if (unlikely(rollover)) {
auto nlp = nl.end();
auto p = lines.end();
assert(lines.size() + rollover < lines.capacity());
lines.resize(lines.size() + rollover);
auto dest = lines.end();
while (--p < --dest) {
if (*p == '\n'+1) {
--*p;
*dest-- = '1';
}
if (*p == '\n') {
*nlp-- = dest + 1;
}
*dest = *p;
}
}
}
}



Compile using g++-10 -std=c++2a -Wall -Wextra -Weffc++ -DNDEBUG -O3 -march=native -fno-exceptions -lpthread. No special arguments or environment are needed when running.

• Neil + Kamila generates fizzbuzz About 50% faster than the kernel generates 0's on my machine (9 vs 6 gib/s, approx), so I wouldn't consider /dev/zero an upper limit. An interesting reference point, maybe. In any case, tried running your new code - optimal when I fake nproc to be 4 - but it just yields results as fast as your previous submission with only 1 core. Jan 22 '21 at 16:42
• That's a shame - unmodified, it runs about 50% faster than my other one here. I really expected to come out ahead on your system too. Jan 22 '21 at 20:11
• I've now spent far longer than I should on this. Time to bow out gracefully. Jan 22 '21 at 20:12

C++

I’m not sure how you did the benchmarking but would appreciate it if you can run my naïve solution through it.

PS, sorry for being late for the party.

#include <vector>
#include <iostream>
#include <unistd.h>

class worker
{
public:
worker(size_t size)
{
workersize = size;
buf = new char[workersize * 40];
}

~worker()
{
delete buf;
}

void run()
{
char thisMod = 0;
char ibMod;
char ib[20];
size_t ibStart;
size_t s;
size_t last_ib = -100;
char* ptr = buf;

for (size_t i = start; i < end; i++)
{
size_t m3 = i % 3;
size_t m5 = i % 5;
if (m3 == 0 && m5 == 0)
{
*ptr++ = 'F';
*ptr++ = 'i';
*ptr++ = 'z';
*ptr++ = 'z';
*ptr++ = 'B';
*ptr++ = 'u';
*ptr++ = 'z';
*ptr++ = 'z';
}
else if (m3 == 0)
{
*ptr++ = 'F';
*ptr++ = 'i';
*ptr++ = 'z';
*ptr++ = 'z';
}
else if (m5 == 0)
{
*ptr++ = 'B';
*ptr++ = 'u';
*ptr++ = 'z';
*ptr++ = 'z';
}
else
{
if (i - last_ib < 10 && ibMod + thisMod < 10)
{
// uses previous ib
for (s = ibStart; s <= 14; s++)
*ptr++ = ib[s];
*ptr++ = ib[s] + thisMod - ibMod;
}
else
{
// calc new ib
size_t x = i;
size_t s = 15;
ibMod = x % 10;
ib[s--] = ibMod + '0';
x /= 10;
while (x > 0)
{
ib[s--] = (x % 10) + '0';
x /= 10;
}
ibStart = ++s;
for (ibStart; s <= 15; s++)
*ptr++ = ib[s];
last_ib = i;
}
}

*ptr++ = '\n';
thisMod++;
if (thisMod > 9)
thisMod = 0;
}
buflen = ptr - buf;
}

size_t start;
size_t end;
size_t workersize;
char* buf;
size_t buflen;
};

{
w->run();
}

int main()
{
size_t workersize = 10000000;

// create workers
std::vector<worker*> workers;
for (size_t i = 0; i < workercount; i++)
workers.push_back(new worker(workersize));

// main loop
size_t cur = 0;
for (;;)
{
// init workers
for (worker* worker : workers)
{
worker->start = cur;
cur += workersize;
worker->end = cur;
}

for (worker* worker : workers)

// write output
for (worker* worker : workers)
write(1, worker->buf, worker->buflen);

// write progress to cerr
std::cerr << cur << "\n";
}
}

• Not bad. Seems to fluctuate between 4GiB/s to 5.5GiB/s. Compiled with g++ a.cpp -lpthread -o Johan -O3. When I run it all my 32 cores are about 20-40% busy Oct 8 '21 at 14:27
• Using clang++ a.cpp -lpthread -o Johanclang -O3 didn't make much of a difference Oct 8 '21 at 14:34
• @OmerTuchfeld, thank you very much for testing!! Can you perhaps run my latest code when you get a chance?
– jdt
Oct 8 '21 at 17:28
• No noticeable difference in throughput Oct 8 '21 at 21:53
• @OmerTuchfeld Oh well, thanks anyway! I think that is as far as I can push it. I got an average of about 4 GB/s on my 6 core i7-9850H with the following TIO and was hoping that running it on your 32-core monster would have made a big difference.
– jdt
Oct 8 '21 at 23:58

There's plenty room for improvement here and I'll try to update when I can. (Most of the optimizations above will also be applicable):

A simple Powershell solution:

$i = 1 #While ($i -gt 0){
While ($i -lt 100000){ # Limiting to 100k for testing if ((($i % 5) -eq 0) -and (($i % 3) -eq 0)){Write-Host "FizzBuzz"} elseif (($i % 3) -eq 0) {Write-Host "Fizz"}
elseif (((($i -split "" | Select-Object -SkipLast 1) | Select-Object -Last 1) - eq 0) -or ((($i -split "" |
Select-Object -SkipLast 1) | Select-Object -Last 1) -eq 5)){Write-Host "Buzz"}
else {Write-Host $i}$i++}

• Tried running it with podman run -it --rm --name powershell --volume $PWD:/root mcr.microsoft.com/powershell /usr/bin/pwsh /root/fizz.ps1 without success: Line | 7 | Select-Object -Last 1) - eq 0) -or ((($i -split "" | | ~ | You must provide a value expression following the '-' operator. Dec 10 '21 at 21:05

Ruby

# frozen_string_literal: true
FMT = "%d\n%d\nFizz\n%d\nBuzz\nFizz\n%d\n%d\nFizz\nBuzz\n%d\nFizz\n%d\n%d\nFizzBuzz"
(1..).each_slice(15) do |slice|
puts format(FMT, slice[0], slice[1], slice[3], slice[6], slice[7], slice[10], slice[12], slice[13])
end


Doesn't seem to be as magical as system languages though. On MacBook Air 1,7 GHz Dual-Core Intel Core i7 8 GB 1600 MHz DDR3 (while watching a twitch stream)

I'm getting from ruby fizzbuzz.rb | pv > /dev/null:

• [44.4MiB/s] for ruby 2.7.4 but seems fuctuating
• [39.3MiB/s] for ruby 3.1-dev but seems to be more stable
• Getting 100MiB/s ruby 3.0.2p107 (2021-07-07 revision 0db68f0233) [x86_64-linux] Dec 9 '21 at 0:07
• I hope this qualifies for the bounty then :) Dec 9 '21 at 13:36
• Sorry that bounty did not come from me Dec 9 '21 at 14:27
• I saw that was not yours. I thought @ksousa wanted dynamic languages. But he rewarded java instead :) Dec 10 '21 at 16:31

This probably won't be the fastest, but I am curious how it performs compared to a naive stdio-based fizzbuzz. I created this to illustrate the approach I suggested in the comments of using an array of fixed-length regions to format the output, and filtering out null padding before printing the output. I didn't do any multithreading, either, this could probably be improved by using multiple buffers and doing the writing on another thread.

This code does nothing to prevent unusual output after reaching ULONG_MAX. It might be able to be made faster by only supporting up to UINT_MAX, which would also allow more data to fit in a given size of intermediate buffer.

// sample outputs [space is null]
//                    1\n
// Fizz                \n
//     Buzz            \n
// FizzBuzz            \n
// 18446744073709551614\n ULLONG_MAX-1 [ULLONG_MAX is FizzBuzz]
#define LEN 21
//#define NUM (200*15)
#define NUM (200*15)

#if defined(_POSIX_C_SOURCE)
#include <unistd.h>
#define WRITE(b,l) write(1,b,l)
#elif defined(_MSC_VER)
#include <io.h>
#define WRITE(b,l) write(1,b,l)
#else
#include <stdio.h>
#define WRITE(b,l) fwrite(b,1,l,stdout)
#endif

char buf[NUM*LEN] = {0}; // 63000
char buf2[NUM*LEN];
int flags[NUM];

int main() {
// pre-populate buffer
for(int i=0; i<NUM; i++) {
char *t = buf+i*LEN;
int f = flags[i] = i%3==0 | ((i%5==0)<<1);
if(f&1) { t[0]='F'; t[1]='i'; t[2]='z'; t[3]='z';}
if(f&2) { t[4]='B'; t[5]='u'; t[6]='z'; t[7]='z';}
t[LEN-1] = '\n';
}
int j=1; // position of current output within buffer
for(unsigned long long i=1;;i++) {
if(!flags[j]) {
char *t = buf+j*LEN+(LEN-2); // position of last digit
unsigned long long k = i;
while(k) {
*t-- = '0'+k%10;
k/=10;
}
}
if(++j == NUM) {
j=0;
char *p=buf;
char *q=buf2;
if(i == NUM-1) p+=LEN; // skip zero, there's probably a better way
while(p < buf+NUM*LEN) {
if(*p) *q++=*p;
p++;
}
WRITE(buf2, q-buf2);
}
}
}

• I'm getting 0.5GiB/s Oct 29 '21 at 21:58
• yeah, once i posted it i realized that per-character branch inside the while loop is probably terrible for the performance, i'd been hoping it was optimized away to a conditional move or something - i can't find a way to get rid of it either. Ah well, it's at least useful as evidence this approach isn't worthwhile. Oct 29 '21 at 22:01

Python

[print('FizzBuzz' if i % 15 == 0 else 'Fizz' if i % 3 == 0 else 'Buzz' if i % 5 == 0 else i) for i in range(10000)]

[print(                   |
'FizzBuzz' if i % 15 == 0 | If the variable i is divisible by 3 and 5 then print 'FizzBuzz'
else 'Fizz' if i % 3 == 0 | If the variable i is divisible by 3 then print 'Fizz'
else 'Buzz' if i % 5 == 0 | If the variable i is divisible by 5 then print 'Buzz'
else i                    | If none of the above is true, then print the variable i
)                         |
for i in range(10000)]    | Repeat the above 10000 times, setting i to 0, and then 1, up to 9999

• Welcome to Code Golf! Per the challenge, "The output must go on forever (or a very high astronomical number) and not halt / change prematurely.", and this terminates after only 10000` iterations (i.e. prematurely) Dec 5 '21 at 21:41