149
\$\begingroup\$

Write the shortest program in your favourite language to interpret a brainfuck program. The program is read from a file. Input and output are standard input and standard output.

  1. Cell size: 8bit unsigned. Overflow is undefined.
  2. Array size: 30000 bytes (not circled)
  3. Bad commands are not part of the input
  4. Comments begin with # and extend to the end of line Comments are everything not in +-.,[]<>
  5. no EOF symbol

A very good test can be found here. It reads a number and then prints the prime numbers up to that number. To prevent link rot, here is a copy of the code:

compute prime numbers
to use type the max number then push Alt 1 0
===================================================================
======================== OUTPUT STRING ============================
===================================================================
>++++++++[<++++++++>-]<++++++++++++++++.[-]
>++++++++++[<++++++++++>-]<++++++++++++++.[-]
>++++++++++[<++++++++++>-]<+++++.[-]
>++++++++++[<++++++++++>-]<+++++++++.[-]
>++++++++++[<++++++++++>-]<+.[-]
>++++++++++[<++++++++++>-]<+++++++++++++++.[-]
>+++++[<+++++>-]<+++++++.[-]
>++++++++++[<++++++++++>-]<+++++++++++++++++.[-]
>++++++++++[<++++++++++>-]<++++++++++++.[-]
>+++++[<+++++>-]<+++++++.[-]
>++++++++++[<++++++++++>-]<++++++++++++++++.[-]
>++++++++++[<++++++++++>-]<+++++++++++.[-]
>+++++++[<+++++++>-]<+++++++++.[-]
>+++++[<+++++>-]<+++++++.[-]

===================================================================
======================== INPUT NUMBER  ============================
===================================================================
+                          cont=1
[
 -                         cont=0
 >,
 ======SUB10======
 ----------

 [                         not 10
  <+>                      cont=1
  =====SUB38======
  ----------
  ----------
  ----------
  --------

  >
  =====MUL10=======
  [>+>+<<-]>>[<<+>>-]<     dup

  >>>+++++++++
  [
   <<<
   [>+>+<<-]>>[<<+>>-]<    dup
   [<<+>>-]
   >>-
  ]
  <<<[-]<
  ======RMOVE1======
  <
  [>+<-]
 ]
 <
]
>>[<<+>>-]<<

===================================================================
======================= PROCESS NUMBER  ===========================
===================================================================

==== ==== ==== ====
numd numu teid teiu
==== ==== ==== ====

>+<-
[
 >+
 ======DUP======
 [>+>+<<-]>>[<<+>>-]<

 >+<--

 >>>>>>>>+<<<<<<<<   isprime=1

 [
  >+

  <-

  =====DUP3=====
  <[>>>+>+<<<<-]>>>>[<<<<+>>>>-]<<<

  =====DUP2=====
  >[>>+>+<<<-]>>>[<<<+>>>-]<<< <


  >>>


  ====DIVIDES=======
  [>+>+<<-]>>[<<+>>-]<   DUP i=div

  <<
  [
    >>>>>+               bool=1
    <<<
    [>+>+<<-]>>[<<+>>-]< DUP
    [>>[-]<<-]           IF i THEN bool=0
    >>
    [                    IF i=0
      <<<<
      [>+>+<<-]>>[<<+>>-]< i=div
      >>>
      -                  bool=0
    ]
    <<<
    -                    DEC i
    <<
    -
  ]

  +>>[<<[-]>>-]<<          
  >[-]<                  CLR div
  =====END DIVIDES====


  [>>>>>>[-]<<<<<<-]     if divides then isprime=0


  <<

  >>[-]>[-]<<<
 ]

 >>>>>>>>
 [
  -
  <<<<<<<[-]<<

  [>>+>+<<<-]>>>[<<<+>>>-]<<<

  >>




  ===================================================================
  ======================== OUTPUT NUMBER  ===========================
  ===================================================================
  [>+<-]>

  [
   ======DUP======
   [>+>+<<-]>>[<<+>>-]<


   ======MOD10====
   >+++++++++<
   [
    >>>+<<              bool= 1
    [>+>[-]<<-]         bool= ten==0
    >[<+>-]             ten = tmp
    >[<<++++++++++>>-]  if ten=0 ten=10
    <<-                 dec ten     
    <-                  dec num
   ]
   +++++++++            num=9
   >[<->-]<             dec num by ten

   =======RROT======
      [>+<-]
   <  [>+<-]
   <  [>+<-]
   >>>[<<<+>>>-]
   <

   =======DIV10========
   >+++++++++<
   [
    >>>+<<                bool= 1
    [>+>[-]<<-]           bool= ten==0
    >[<+>-]               ten = tmp
    >[<<++++++++++>>>+<-] if ten=0 ten=10  inc div
    <<-                   dec ten     
    <-                    dec num
   ]
   >>>>[<<<<+>>>>-]<<<<   copy div to num
   >[-]<                  clear ten

   =======INC1=========
   <+>
  ]

  <
  [
   =======MOVER=========
   [>+<-]

   =======ADD48========
   +++++++[<+++++++>-]<->

   =======PUTC=======
   <.[-]>

   ======MOVEL2========
   >[<<+>>-]<

   <-
  ]

  >++++[<++++++++>-]<.[-]

  ===================================================================
  =========================== END FOR ===============================
  ===================================================================


  >>>>>>>
 ]
 <<<<<<<<



 >[-]<
  [-]
 <<-
]

======LF========

++++++++++.[-]
@

Example run:

$ python2 bf.py PRIME.BF 
Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 
\$\endgroup\$
27
  • 9
    \$\begingroup\$ You should clarify about 1) size of memory 2) is memory circled 4) maybe any other details \$\endgroup\$
    – Nakilon
    Jan 28, 2011 at 1:37
  • 3
    \$\begingroup\$ I wonder if there should be two categories: Those programs that use eval (or shell out to compile) -- and those that don't. \$\endgroup\$ Feb 15, 2011 at 7:52
  • 47
    \$\begingroup\$ I'd love to see someone answer this in brainfuck. \$\endgroup\$
    – Hannesh
    Mar 14, 2011 at 19:15
  • 8
    \$\begingroup\$ What does "no EOF symbol" mean? That the cell value remains unchanged when trying , on EOF? Or that it's up to us to choose a value when trying , on EOF? Or is EOF undefined behaviour altogether? \$\endgroup\$ Apr 1, 2016 at 14:07
  • 4
    \$\begingroup\$ Likewise, what should happen when someone tries to leave the 30k cells to either side? Should the tape head remain in place or is this undefined behaviour? \$\endgroup\$ Apr 1, 2016 at 14:09

78 Answers 78

84
\$\begingroup\$

Python (no eval), 317 bytes

from sys import*
def f(u,c,k):
 while(c[1]>=k)*u:
  j,u='[]<>+-,.'.find(u[0]),u[1:];b=(j>=0)*(1-j%2*2);c[1]+=b*(j<2)
  while b*c[c[0]]and j<1:f(u,c,k+1);c[1]+=1
  b*=c[1]==k;c[[0,c[0],2][j/2-1]]+=b
  if(j==6)*b:c[c[0]]=ord(stdin.read(1))
  if(j>6)*b:stdout.write(chr(c[c[0]]))
f(open(argv[1]).read(),[-1]+[0]*30003,0)
\$\endgroup\$
4
  • 18
    \$\begingroup\$ That is one beautiful piece of noise, sir \$\endgroup\$
    – globby
    Dec 6, 2014 at 1:44
  • 2
    \$\begingroup\$ -1 byte if you replace while b*c[c[0]]and j<1 with while b*c[c[0]]*(j<1) \$\endgroup\$ Jun 24, 2019 at 7:58
  • 2
    \$\begingroup\$ -2 bytes if you remove the whitespace as well: while(j<1)*b*c[c[0]] \$\endgroup\$
    – movatica
    Aug 11, 2020 at 21:34
  • 1
    \$\begingroup\$ -1 You said a naughty word in your function definition. \$\endgroup\$ Sep 4, 2023 at 16:32
79
\$\begingroup\$

brainfuck, 843 691 bytes

Edit: decided to revisit this and found a surprising number of ways to golf off bytes

>>>,[>++++[-<-------->]<-[>+<<]>[----------[>]>[<+<+>>>>]<<<-[>]>[<+<+>>>>]<<<-[>]>[<+<+>>>>]<<<-[>]>[<-<+++>>>>]<<<--------------[>]>[<++<+>>>>]<<<--[>]>[<-<+++++++>>+>>]<<++++[-<------>]+<+[>]>[<++<+>>>>]<<<--[>]>[<<+>>>>]<<-<[+]<[>]>,>]<]<-[<]>[-[<<]>[<+[>]>>[<+[<<[<]<<-[>>]<[>>>>[>]>+<<[<]<]<-[>>]<[>>>>[>]>-<<[<]<]<++[->>+<<]>>[>]>]]<<<[<]>-<]>-[<<]>[<++[>]>+>[<-]<[<<[<]>[-<<+>>]>--[<<]>[[>]>+<<[<]<]>+[<<]>[[>]>-<<[<]<]>+[>]>]<<[<]>--<]>-[<<]>[[>]>>.<<<[<]<]>-[<<]>[[>]>>-<<<[<]<]>-[<<]>[[>]>>,<<<[<]<]>-[<<]>[[>]>>+<<<[<]<]>-[<<]>[[>]>>>[>>]>[<<<[<<]<+>>>[>>]>-]>[-]<<+[<[->>+<<]<]<[->>+<<]<[<]<]>-[<<]>[[>]>-[+>[-<<+>>]>]+<<[-]+[-<<]<[->>>[>>]>+<<<[<<]<]<[<]<]<++++++++>>[+<<->>]>]

This takes input in the form code!input where the !input is optional. It also simulates negative cells without using negative cells itself and can store up to (30000-(length of code+6))/2 cells.

Try it online!

\$\endgroup\$
5
  • 5
    \$\begingroup\$ Just to make sure I got this right, if I ran this program with this program I could nest it 5 levels deep and still handling code-inputs of length 262. \$\endgroup\$ Oct 1, 2019 at 21:46
  • 1
    \$\begingroup\$ @Draco18s I suspect you'd run out the 30000 cells before that, since the size of each nested interpreter increases exponentially. I think you'd get 2, maybe 3 levels deep \$\endgroup\$
    – Jo King
    Oct 1, 2019 at 23:03
  • 2
    \$\begingroup\$ Even 3 deep would be hilariously silly. \$\endgroup\$ Oct 2, 2019 at 2:34
  • \$\begingroup\$ It seems to only work two layers deep, 3 just makes it hang \$\endgroup\$
    – the-cobalt
    Sep 5, 2020 at 1:47
  • 5
    \$\begingroup\$ @the-cobalt It's not hanging, it's just taking an extremely long time. On copy.sh, it takes a mere 50 minutes to execute that program \$\endgroup\$
    – Jo King
    Sep 5, 2020 at 6:54
54
\$\begingroup\$

16 bit 8086 machine code: 168 bytes

Here's the base64 encoded version, convert and save as 'bf.com' and run from Windows command prompt: 'bf progname'

gMYQUoDGEFKzgI1XAgIfiEcBtD3NIR8HcmOL2LQ/i88z0s0hcleL2DPA86sz/zP2/sU783NHrL0I
AGgyAU14DTqGmAF194qOoAH/4UfDJv4Fwyb+DcO0AiaKFc0hw7QBzSGqT8MmODV1+jPtO/NzDaw8
W3UBRTxddfJNee/DJjg1dPoz7U509YpE/zxddQFFPFt18U157sM+PCstLixbXUxjTlJWXmV+

EDIT

Here's some assembler (A86 style) to create the executable (I had to reverse engineer this as I'd misplaced the original source!)

    add dh,10h                              
    push dx                                 
    add dh,10h                              
    push dx                                 
    mov bl,80h                              
    lea dx,[bx+2]                         
    add bl,[bx]                            
    mov [bx+1],al                         
    mov ah,3dh                              
    int 21h                                 
    pop ds                                 
    pop es                                 
    jb ret                               
    mov bx,ax                              
    mov ah,3fh                              
    mov cx,di                              
    xor dx,dx                              
    int 21h                                 
    jb ret                               
    mov bx,ax                              
    xor ax,ax                              
    repz stosw                                     
    xor di,di                              
    xor si,si                              
    inc ch                                 
program_loop:
    cmp si,bx                              
    jnb ret                               
    lodsb                                    
    mov bp,8                            
    push program_loop
symbol_search:                       
    dec bp                                 
    js ret
    cmp al,[bp+symbols]
    jnz symbol_search
    mov cl,[bp+instructions]
    jmp cx                                 
forward:
    inc di                                 
    ret                                    
increment:
    inc b es:[di]                      
    ret                                    
decrement:
    dec b es:[di]                      
    ret                                    
output:
    mov ah,2                              
    mov dl,es:[di]                            
    int 21h                                 
    ret                                    
input:
    mov ah,1                              
    int 21h                                 
    stosb                                    
backward:
    dec di                                 
    ret                                    
jumpforwardifzero:
    cmp es:[di],dh                            
    jnz ret                               
    xor bp,bp
l1: cmp si,bx                              
    jnb ret
    lodsb                                    
    cmp al,'['                              
    jnz l2
    inc bp
l2: cmp al,']'                              
    jnz l1
    dec bp                                 
    jns l1
    ret                                    
jumpbackwardifnotzero:
    cmp es:[di],dh                            
    jz  ret
    xor bp,bp
l3: dec si                                 
    jz  ret
    mov al,[si-1]                         
    cmp al,']'
    jnz l4
    inc bp  
l4: cmp al,'['                              
    jnz l3
    dec bp                                 
    jns l3
    ret                                    
symbols:
    db '><+-.,[]'
instructions:
    db forward and 255
    db backward and 255
    db increment and 255
    db decrement and 255
    db output and 255
    db input and 255
    db jumpforwardifzero and 255
    db jumpbackwardifnotzero and 255
\$\endgroup\$
2
  • \$\begingroup\$ I've added a source code version of the program. I've just noticed that non-bf characters cause the program to exit rather than be ignored. Easy to fix that and I'll leave it as an exercise for people to do that themselves. \$\endgroup\$
    – Skizz
    Aug 14, 2012 at 13:18
  • 2
    \$\begingroup\$ I remember I got the Linux ELF version 166 bytes, 10 years ago, here muppetlabs.com/~breadbox/software/tiny \$\endgroup\$
    – Emmanuel
    Jul 28, 2014 at 17:34
48
\$\begingroup\$

Perl, 120 138

%c=qw(> $p++ < $p-- + D++ - D-- [ while(D){ ] } . print+chrD , D=ord(getc));
$/=$,;$_=<>;s/./$c{$&};/g;s[D]'$b[$p]'g;eval

This runs hello.bf and primes.bf flawlessly:

$ perl bf.pl hello.bf
Hello World!
$ perl bf.pl prime.bf
Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97

Initialization: The opcode to Perl translation table is stored in %c. The readable form looks like this:

%c=(
  '>' => '$p++',
  '<' => '$p--',
  '+' => '$b[$p]++',
  '-' => '$b[$p]--',
  '[' => 'while($b[$p]){',
  ']' => '}',
  '.' => 'print chr$b[$p]',
  ',' => '$b[$p]=ord(getc)',
);

Step 1: Slurp program input to $_ and transform it to Perl code using the translation table. Comments are automatically stripped (replaced with undef) in this step.

Step 2: Uncompress all $b[$p] occurrences

Step 3: Launch the program using eval.

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Just use Perl's qw syntax to define %c directly -- good for 7 fewer chars (you'll have to say print+chr$b[$p] and ord(getc), though) \$\endgroup\$
    – mob
    Sep 11, 2012 at 23:36
  • \$\begingroup\$ I count 18 saved… thanks! (updating in a minute) \$\endgroup\$
    – J B
    Sep 12, 2012 at 8:19
  • \$\begingroup\$ The %c table is declared and defined in the first line; its characters are perfectly accounted for. \$\endgroup\$
    – J B
    Sep 19, 2012 at 13:46
  • \$\begingroup\$ Someone called "Grimy" did it on 105 bytes when competing at this competition (web.archive.org/web/20131113171000/http://codegolf.com/…) How to reach Grimy ? Do you know him ? \$\endgroup\$
    – YOGO
    Jun 15, 2022 at 23:06
38
+100
\$\begingroup\$

Vim, 809 580 bytes

-229 bytes because shenanigans and balderdash.

qq<C-v><C-v><C-v>q<Esc>^mbO<C-r>q
elsei "<C-r>q<C-r>c"=="<Esc>^"xc$<C-r>q
norm <Esc>^"zc$en<C-r>q
<Esc>^"yd$i`b"cyl:if "<C-v><C-r>c"==">"<C-r>z@f<C-r>x<"<C-r>z@b<C-r>x+"<C-r>z@i<C-r>x-"<C-r>z@d<C-r>x,"<C-r>z@p<C-r>x."<C-r>z@o<C-r>x["<C-r>z@s<C-r>x]"<C-r>z@l<C-v>
<C-r>y`blmb@e<Esc>^"ec$`b^mbjj30000ddk30000o0<C-v><Esc>jjmo`lk$d`ox`bjjmp`ljmi`b<Esc>^"rc$`p:if line(".")<30002<C-r>zj<C-v>
<C-r>ymp<Esc>^"fc$`p:if line(".")>3<C-r>zk<C-v>
<C-r>ymp<Esc>^"bc$`p<C-v><C-a>Y:if <C-v><C-r>"<C-v><C-h>>255<C-r>z<C-v><C-x><C-v>
<C-r>y<Esc>^"ic$`p<C-v><C-x>Y:if <C-v><C-r>"<C-v><C-h><0<C-r>z<C-v><C-a><C-v>
<C-r>y<Esc>^"dc$`i"qx`p:if "<C-v><C-r>""!="<C-r>q<Esc>"<C-r>z3xi<C-r>q<C-r>=char2nr("<C-r>q<C-r>q")<C-r>q
<C-r>q<Esc><C-v>
else<C-r>z`i3i<C-v><C-v><Esc>a<C-v><Esc><C-r>q<Esc><C-v>
<C-r>y<Esc>^"pc$`py$`oA<C-v><C-r>=nr2char(<C-v><C-r>")<C-v>
<C-v><Esc>$mo<Esc>^"oc$`py$`b:if <C-v><C-r>"==0<C-r>z%mb<C-v>
<C-r>y<Esc>^"sc$`py$`b:if <C-v><C-r>"!=0<C-r>z%mb<C-v>
<C-r>y<Esc>^"ld$ddo<Esc>90i-<Esc>30000o0<Esc>o<Esc>90i-<Esc>o<Esc>moo<Esc>ml90i-<Esc>o<C-v><Esc><Esc>mio<Esc>90i-<Esc>`bjjmp

Try it online!

More than just an interpreter, this sets up an environment for programming in brainfuck with the following format:

brainfuck program
-------------------------------------------------------------------
list
of
30000
memory
cells
-------------------------------------------------------------------
program output
-------------------------------------------------------------------
program input ^[
-------------------------------------------------------------------

It also sets up some macros:

  • @f - > moves the MP forward
  • @b - < moves the MP backward
  • @i - + increments the cell at the MP
  • @d - - decrements the cell at the MP
  • @p - , input a character and store it in the cell at the MP
  • @o - . output the signified by the cell at the MP
  • @s - [ start while loop
  • @l - ] end while loop
  • @e - Execute. Recursively get the command at the IP and call the corresponding macro if it is a valid command, then increment the IP.
  • @r - Reset. Clear the output, reset all memory cells, and move the IP back to the beginning of the program.

If using the TIO link, the input box is where the program goes to be executed. To pass input to the brainfuck program, add this to the footer before anything else: `iiYour Input Here<Esc>. Additionally, I added gg30003dd`l3dd to the footer to strip away everything except the brainfuck program's output, so you can remove that from the footer to see the entire result.

If using Vim, copy the TIO code and paste it into Vim to set up the environment. From there, you can use `b to jump to the brainfuck program, `p to jump to the MP, `o to jump to the output, and `i to jump to the input.

A few caveats:

  • The environment does not support newlines in the program, so the brainfuck program must be a single line.
  • The input must be terminated with <Esc>. The setup code automatically puts the <Esc> in the input section, so just make sure that the input goes before the ^[.
  • To "read the program from a file", as it were, you can use Vim to open a file containing the brainfuck program and paste this code to set everything up and then execute it, as long as the program is already in a single line.
  • It's kinda slow. The test program for finding primes took ~2 minutes to finish with n=2. It took ~10 minutes with n=7. I didn't even bother trying anything higher. Nevertheless, it works.

Note: This is the second version, and is golfed much further than the original, but I could almost certainly get it even shorter by defining the macros as macros instead of deleting into named registers. However, that will require pretty much rewriting this entire program, which I just golfed into illegibility, so I don't feel like doing it right now, but I might do it some time later.

\$\endgroup\$
2
  • 11
    \$\begingroup\$ what in the world.. \$\endgroup\$
    – Razetime
    Apr 20, 2021 at 14:32
  • 9
    \$\begingroup\$ I've been delaying learning vim for quite some time now, I have to learn vim after seeing this answer \$\endgroup\$
    – user100752
    Jun 20, 2021 at 16:54
29
\$\begingroup\$

Ruby 1.8.7, 188 185 149 147 characters

eval"a=[i=0]*3e4;"+$<.bytes.map{|b|{?.,"putc a[i]",?,,"a[i]=getc",?[,"while a[i]>0",?],"end",?<,"i-=1",?>,"i+=1",?+,"a[i]+=1",?-,"a[i]-=1"}[b]}*";"

Somewhat readable version:

code = "a = [0] * 3e4; i = 0;"
more_code ARGF.bytes.map {|b|
  replacements = {
    ?. => "putc a[i]",
    ?, => "a[i] = getc",
    ?[ => "while a[i] > 0 do",
    ?] => "end",
    ?< => "i -= 1",
    ?> => "i += 1",
    ?+ =>"a[i]+=1",
    ?- =>"a[i]-=1"
  }
  replacements[b]
}.join(";")
eval code+more_code

As you see I shamelessly stole your idea of translating to the host language and then using eval to run it.

\$\endgroup\$
13
  • \$\begingroup\$ You can shave off a byte byte comparing to zero >0 rather than testing equality: !=0. The specs say unsigned, and overflow is undefined. \$\endgroup\$ Feb 6, 2011 at 5:32
  • \$\begingroup\$ 3e4 will also work as opposed to 30000 \$\endgroup\$ Feb 6, 2011 at 5:36
  • \$\begingroup\$ @Charlie: Thanks. Though to be fair it didn't say "unsigned" when I wrote the code. I honestly didn't know that you could write 3e4 though. That's a very good point and good to know. \$\endgroup\$
    – sepp2k
    Feb 6, 2011 at 5:40
  • \$\begingroup\$ File.read($*.pop).bytes -> $<.bytes should work too \$\endgroup\$ Feb 12, 2011 at 22:31
  • 1
    \$\begingroup\$ Ruby 1.8.7 has an even shorter syntax to build a literal hash: {?a,"foo"}, which is equivalent to {?a=>"foo"}. And testing here shows that you actually can replace File.read($*.pop).bytes with $< without any problems. Also inlining everything to something like eval"a[0]..."+$<.bytes.map{?.,"putc a[i]",...}*";" shortens the solution by another few characters. \$\endgroup\$
    – Ventero
    Feb 14, 2011 at 15:58
29
\$\begingroup\$

Binary Lambda Calculus 112

The program shown in the hex dump below

00000000  44 51 a1 01 84 55 d5 02  b7 70 30 22 ff 32 f0 00  |DQ...U...p0".2..|
00000010  bf f9 85 7f 5e e1 6f 95  7f 7d ee c0 e5 54 68 00  |....^.o..}...Th.|
00000020  58 55 fd fb e0 45 57 fd  eb fb f0 b6 f0 2f d6 07  |XU...EW....../..|
00000030  e1 6f 73 d7 f1 14 bc c0  0b ff 2e 1f a1 6f 66 17  |.os..........of.|
00000040  e8 5b ef 2f cf ff 13 ff  e1 ca 34 20 0a c8 d0 0b  |.[./......4 ....|
00000050  99 ee 1f e5 ff 7f 5a 6a  1f ff 0f ff 87 9d 04 d0  |......Zj........|
00000060  ab 00 05 db 23 40 b7 3b  28 cc c0 b0 6c 0e 74 10  |....#@.;(...l.t.|
00000070

expects its input to consist of a Brainfuck program (looking only at bits 0,1,4 to distinguish among ,-.+<>][ ) followed by a ], followed by the input for the Brainfuck program.

Save the above hex dump with xxd -r > bf.Blc

Grab a blc interpreter from https://tromp.github.io/cl/cl.html

cc -O2 -DM=0x100000 -m32 -std=c99 uni.c -o uni
echo -n "++++++++++[>+++++++>++++++++++>+++>+<<<<-]>++.>+.+++++++..+++.>++.<<+++++++++++++++.>.+++.------.--------.>+.>.]" > hw.bf
cat bf.Blc hw.bf | ./uni

Hello World!

\$\endgroup\$
3
  • 1
    \$\begingroup\$ Why does this even exist? Apparently, it even exists in the realm of research. O.o \$\endgroup\$
    – Claudia
    Dec 4, 2014 at 4:11
  • \$\begingroup\$ So this wouldn't work with commented brainfuck programs? \$\endgroup\$
    – kamoroso94
    Feb 15, 2018 at 20:05
  • \$\begingroup\$ No, not without stripping out the comments first. \$\endgroup\$
    – John Tromp
    Feb 16, 2018 at 22:27
18
\$\begingroup\$

Retina 0.8.2, 386 391 386 bytes

Code contains unprintable NUL (0x00) characters. It's also not super golfed yet, because it's already really slow, and if I golf it more, I don't know how long it'd take to finish. Appears to time out on the prime-finding sample.

There may be bugs in the online interpreter or in my program (leading new lines don't show in the output?).

Takes input like <code>│<input>. No, that is not a pipe (|). It's the Unicode character U+2502. The code also uses the Unicode characters ÿ▶◀├║. Unicode characters are used in order to support input of all ASCII characters. Therefore, these characters need to be separated from the code by a non-ASCII character.

Try it online

s`^.*
▶$0├║▶
s{`(▶>.*║.*)▶(.)(.?)
$1$2▶$3
▶$
▶
║▶
║▶
(▶<.*║.*)(.)▶
$1▶$2
T`ÿ-`o`(?<=▶\+.*║.*▶).
^\+

T`-ÿ`ÿo`(?<=▶-.*║.*▶).
^-

(▶\..*├.*)(║.*▶)(.)
$1$3$2$3
(▶,.*│)(.?)(.*├.*▶).
$1$3$2
▶\[(.*║.*▶)
[▶▶${1}
{`(▶▶+)([^[\]]*)\[
$2[$1▶
}`▶(▶+)([^[\]]*)\]
$2]$1
r`([[\]]*)▶\](.*║.*▶[^])
$1◀◀]$2
r{`\[([^[\]]*)(◀+)◀
$2[$1
}`\]([^[\]]*)(◀◀+)
$2◀]$1
◀
▶
}`▶([^│])(.*║)
$1▶$2
s\`.*├|║.*

Note there is a trailing newline there.

Brief Explanation:

Zeros 0x00 are used for the tape, which is infinite. The first replacement sets up the interpreter in the form ▶<code>│<input>├<output>║▶<tape>, where the first is the pointer for the code, and the second one is the pointer for the tape.

ÿ is 0xFF (255), which is used for Transliteration (used to implement + and -) to wrap the cells back around to zero.

is only used for readability (in case the program is stopped in the middle or you want to see the program mid-execution). Otherwise, you couldn't tell which way the pointer was moving.

Commented Code:

s`^.*                       # Initialize
▶$0├║▶
s{`(▶>.*║.*)▶(.)(.?)        # >
$1$2▶$3
▶$
▶
║▶                          # <
║▶
(▶<.*║.*)(.)▶
$1▶$2
T`ÿ-`o`(?<=▶\+.*║.*▶).      # +
^\+

T`-ÿ`ÿo`(?<=▶-.*║.*▶).      # -
^-

(▶\..*├.*)(║.*▶)(.)         # .
$1$3$2$3
(▶,.*│)(.?)(.*├.*▶).        # ,
$1$3$2
▶\[(.*║.*▶)                 # [
[▶▶${1}
{`(▶▶+)([^[\]]*)\[
$2[$1▶
}`▶(▶+)([^[\]]*)\]
$2]$1
r`([[\]]*)▶\](.*║.*▶[^])    # ]
$1◀◀]$2
r{`\[([^[\]]*)(◀+)◀
$2[$1
}`\]([^[\]]*)(◀◀+)
$2◀]$1
◀
▶
}`▶([^│])(.*║)              # next instruction
$1▶$2
s\`.*├|║.*                  # print output

Click here for the code with zeros in place of null bytes. Any occurrences of $0 should not be replaced with nulls.

Edit: Now supports empty input and suppresses trailing newline.

Infinite output is now supported. (403 bytes)

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1
  • \$\begingroup\$ I kind of wish that I'd placed the <code> and the <tape> next to each other (though it'd be more characters) so that transitioning to an SMBF interpreter would be easier, if I ever decide to do that. \$\endgroup\$
    – mbomb007
    Jul 20, 2016 at 16:43
15
\$\begingroup\$

Conveyor, 953

This might be the most beautiful code you will ever see:

0

:I\1\@p
>#====)
^#====<
PP0
P<=======================<
00t:)01t1  a:P:P:P:P:P:P:^
>===========">">2>">2>">"^
^           +^-^5^ ^5^]^.^
^           "^"^*^"^*^"^"^
^           -^-^6^-^6^-^-^
^           #^#^*^#^*^#^#^
^           P P -^P )^P P
^           P P #^P )^P P
^t1\)t0:))t01   P   -^  1
^===========<   P   #^  0
^  t1\(t0:))t01     P   t
^=============<     P   )
^         t11(t01   0 0 )
^===============<. t P 10
^                 FT#T#=<
^=================< P 
^             t11)t01 
^===================< 10t))0tP00t:(01t(1a:P:
^                     >=====#=>==========">"
^                             ^          ]^[
^                           P ^          "^"
^===========================<=^#=====<   -^-
                            ^==<     ^ PP#^#=
                                     ^===PTPT<
                                     ^  )P P
                                     ^=<=< (
                                       ^===<
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2
  • 12
    \$\begingroup\$ Could you add an explanation and a link to an implementation? I want to understand the beauty. ;) \$\endgroup\$
    – DLosc
    May 12, 2015 at 22:47
  • 3
    \$\begingroup\$ Well, I'm currently developing it, there is a compiler and a very bad explanation at github.com/loovjo/Conveyor. The source is pretty readable if you want to understand it. \$\endgroup\$
    – xenia
    May 13, 2015 at 8:30
14
\$\begingroup\$

Python 275 248 255

I decided to give it a try.

import sys
i=0
b=[0]*30000
t=''
for e in open(sys.argv[1]).read():
 t+=' '*i+['i+=1','i-=1','b[i]+=1','b[i]-=1','sys.stdout.write(chr(b[i]))','b[i]=ord(sys.stdin.read(1))','while b[i]:','pass','']['><+-.,['.find(e)]+'\n'
 i+=(92-ord(e))*(e in'][')
exec t 
\$\endgroup\$
6
  • 15
    \$\begingroup\$ Neat, you are generating python source code using brainfuck. \$\endgroup\$
    – user11
    Jan 30, 2011 at 2:00
  • 1
    \$\begingroup\$ You may strip 1 char, "import sys as s" and replace "sys" to "s" in the rest \$\endgroup\$
    – YOU
    Feb 6, 2011 at 6:52
  • \$\begingroup\$ Note that this is actually 247 chars. (See the nasty space after exec t?). If you use S.Mark's tip and also make the whole for cycle into one line, you can shrink this to 243 chars. \$\endgroup\$ Apr 4, 2011 at 15:18
  • 2
    \$\begingroup\$ I wrote a 227 character version of this in python myself, and there are a few tricks you could use to shorten your code. Rather than storing the number of indents in i, and incrementing/decrementing i, you can store the indent itself, which takes exactly as many characters to update (i=i[e==']':]+' '[e!='[':]), and saves you from doing i*' ', and lets you do i=t=''. Rather than storing '' as the last item in the array, you can keep pass as the last one, which translates any non-operation into pass, which works fine. Everything else I did has been suggested already. \$\endgroup\$
    – Strigoides
    Oct 20, 2012 at 5:42
  • 13
    \$\begingroup\$ I'd +1, but many improvements have been suggested and not implemented. \$\endgroup\$
    – mbomb007
    Nov 15, 2016 at 17:01
14
\$\begingroup\$

TypeScript Types, 1807 1771 bytes

//@ts-ignore
type B<X="\0\x01\x02\x03\x04\x05\x06\x07\b\t\n\v\f\r\x0E\x0F\x10\x11\x12\x13\x14\x15\x16\x17\x18\x19\x1A\x1B\x1C\x1D\x1E\x1F !\"#$%&'()*+,-./0123456789:;<=>?@ABCDEFGHIJKLMNOPQRSTUVWXYZ[\\]^_`abcdefghijklmnopqrstuvwxyz{|}~\x7F\x80\x81\x82\x83\x84\x85\x86\x87\x88\x89\x8A\x8B\x8C\x8D\x8E\x8F\x90\x91\x92\x93\x94\x95\x96\x97\x98\x99\x9A\x9B\x9C\x9D\x9E\x9F\xa0\xa1\xa2\xa3\xa4\xa5\xa6\xa7\xa8\xa9\xaa\xab\xac\xad\xae\xaf\xb0\xb1\xb2\xb3\xb4\xb5\xb6\xb7\xb8\xb9\xba\xbb\xbc\xbd\xbe\xbf\xc0\xc1\xc2\xc3\xc4\xc5\xc6\xc7\xc8\xc9\xca\xcb\xcc\xcd\xce\xcf\xd0\xd1\xd2\xd3\xd4\xd5\xd6\xd7\xd8\xd9\xda\xdb\xdc\xdd\xde\xdf\xe0\xe1\xe2\xe3\xe4\xe5\xe6\xe7\xe8\xe9\xea\xeb\xec\xed\xee\xef\xf0\xf1\xf2\xf3\xf4\xf5\xf6\xf7\xf8\xf9\xfa\xfb\xfc\xfd\xfe\xff",a=[]>=X extends`${infer A}${infer C}`?B<C,[...a,A]>:a;type C={[K in keyof B&`${number}`as B[K]]:D<K>};type D<T,A=[]>=T extends`${A["length"]}`?A["length"]:D<T,[0,...A]>;type E<A=[]>=255 extends A["length"]?[...A,0]:E<[...A,[...A,0]["length"]]>;type F=[255,0,...E];type G<L,D=never>=L extends[infer A,{}]?A:D;type H<L,D=never>=L extends[{},infer B]?B:D;type M<a,b,c=0,d="",e=0,f=0,g=0,h=0,i=[],>=a extends`${infer j}${infer a}`?h extends 0?j extends"+"?M<a,b,c,d,e,E<[]>[f],g>:j extends"-"?M<a,b,c,d,e,F[f],g>:j extends"<"?e extends 0?M<a,b,c,d,e,f,g>:M<a,b,c,d,H<e>,G<e>,[f,g]>:j extends">"?M<a,b,c,d,[f,e],G<g,0>,H<g,[]>>:j extends"["?f extends 0?M<a,b,c,d,e,f,g,1>:M<a,b,[a,c],d,e,f,g>:j extends"]"?f extends 0?M<a,b,H<c>,d,e,f,g>:M<G<c>,b,c,d,e,f,g>:j extends","?b extends`${infer k}${infer b}`?M<a,b,c,d,e,k extends keyof j?j[k]:0,g>:M<a,b,c,d,e,0,g>:j extends"."?M<a,b,c,`${d}${B[f]}`,e,f,g>:M<a,b,c,d,e,f,g>:j extends"]"?M<a,b,c,d,e,f,g,G<i,0>,H<i,[]>>:j extends"["?M<a,b,c,d,e,f,g,1,[1,i]>:M<a,b,c,d,e,f,g,h,i>:d

Try it online!

Ungolfed / Explanation

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1
  • 10
    \$\begingroup\$ How does this work? \$\endgroup\$
    – emanresu A
    Nov 28, 2021 at 18:35
14
\$\begingroup\$

C (gcc) Linux x86_64, 884 621 525 487 439 383 358 347 bytes

*z,*mmap();d[7500];h;(*p)();*j(char*a){char*t=a,*n,c,q=0;for(;read(h,&c,!q);t=c==47?n=j(t+9),z=mempcpy(t,L"\xf003e80Ƅ",5),*z=n-t-9,n:c==49?q=*t++=233,z=t,*z=a-13-t,z+1:stpcpy(t,c-18?c-16?~c?c-1?c-2?c?t:"1\xc0P_\xF\5":"RXR_\xF\5":L"໾":L"۾":L"컿":L"웿"))c-=44;return t;}main(P,g)int**g;{*j(p=mmap(0,1<<20,6,34,h=open(g[1],0),0))=195;p(0,d,1);}

Try it online!

This is a JIT that compiles BF code into x86_64 machine language at runtime. This performs a straight translation so commonly occurring sequences such as >>>, <<<, +++ and --- aren't coalesced into faster instructions.

Less golfed version:

// size of data area
*z,c,*mmap();d[7500];h;(*p)();
// recursive function translates BF commands to x86_64 instructions
*j(char*a){
  char*t=a,*n,q=0;
  for(;read(h,&c,!q);)
    c-=44,
    t=c==47? // [
        // cmpb $0x0,(%rsi)
        // je n-t-9
        n=j(t+9),
        z=mempcpy(t,L"\xf003e80Ƅ",5),
        *z=n-t-9,
        n
      :
        c==49? // ]
          // jmp a-13-t
          q=*t++=233,
          z=t,
          *z=a-13-t,
          z+1
        :
          stpcpy(t,c-18? // >
                     c-16? // <
                       ~c? // +
                         c-1? // -
                           c-2? // .
                             c? // ,
                               ""
                             :
                               // xor %eax,%eax
                               // push %rax
                               // pop %rdi
                               // syscall
                               "1\xc0P_\xF\5"
                           :
                             // push %rdx
                             // pop %rax
                             // push %rdx
                             // pop %rdi
                             // syscall
                             "RXR_\xF\5"
                         :
                           // decb (%rsi)
                           L"໾"
                       :
                         // incb (%rsi)
                         L"۾"
                     :
                       // dec %esi
                       L"컿"
                   :
                     // inc %esi
                     L"웿");
  return t;
}
main(P,g)int**g;{
  // allocate text (executable) memory and mark as executable
  p=mmap(0,1<<20,6,34,h=open(g[1],0),0);
  // run JIT, set %rdx=1 and call code like a function
  p(*j(p)=195,d,1);
}
\$\endgroup\$
1
  • 1
    \$\begingroup\$ save 5 bytes with a=t; instead of return t; \$\endgroup\$
    – jdt
    Sep 21, 2023 at 11:50
13
\$\begingroup\$

Haskell, 457 413 characters

import IO
import System
z=return
'>'#(c,(l,d:r))=z(d,(c:l,r))
'<'#(d,(c:l,r))=z(c,(l,d:r))
'+'#(c,m)=z(succ c,m)
'-'#(c,m)=z(pred c,m)
'.'#t@(c,_)=putChar c>>hFlush stdout>>z t
','#(_,m)=getChar>>=(\c->z(c,m))
_#t=z t
_%t@('\0',_)=z t
i%t=i t>>=(i%)
b('[':r)=k$b r
b(']':r)=(z,r)
b(c:r)=f(c#)$b r
b[]=(z,[])
f j(i,r)=(\t->j t>>=i,r)
k(i,r)=f(i%)$b r
main=getArgs>>=readFile.head>>=($('\0',("",repeat '\0'))).fst.b

This code "compiles" the BF program into an IO action of the form State -> IO State the state is a zipper on an infinite string.

Sad that I had to expend 29 characters to turn buffering off. Without those, it works, but you don't see the prompts before you have to type input. The compiler itself (b, f, and k) is just 99 characters, the runtime (# and %) is 216. The driver w/initial state another 32.

>ghc -O3 --make BF.hs 
[1 of 1] Compiling Main             ( BF.hs, BF.o )
Linking BF ...

>./BF HELLO.BF 
Hello World!

>./BF PRIME.BF 
Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 

update 2011-02-15: Incorporated J B's suggestions, did a little renaming, and tightened up main

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1
  • 1
    \$\begingroup\$ You should be able to get the buffering from just IO, and the arguments from just System (-19). The buffering issue bothers me as well, as the spec doesn't really mention it and the top-voted answer doesn't even do I/O. If you must keep it, it's probably shorter to hFlush after each write than change the global buffering mode (-34+15). \$\endgroup\$
    – J B
    Feb 15, 2011 at 10:49
11
\$\begingroup\$

Brainfuck, 948 bytes

Well, that took a while. I golfed a Brainfuck self-interpreter by ... not me.

->->>>-[,+>+<[->-]>[->]<+<-------------------------------------[+++++++++++++++++++++++++++++++++++++>-]>[->]<<[>++++++++[-<----->]<---[-[-[-[--------------[--[>+++++++[-<---->]<-[--[[+]->]<+[->++>]->]<+[->+>]->]<+[->+++++>]->]<+[->++++++>]->]<+[->+++++++>]->]<+[->++++>]->]<+[->++++++++>]->]<+[->+++>]->]+<+[->->]>[-<->]<]>>->>-<<<<<+++[<]>[-[-[-[-[-[-[-[-<<++++++++>>>[>]>>>>+[->>+]->,<<<+[-<<+]-<<<[<]<]>[<<<+++++++>>>[>]>>>>+[->>+]->.<<<+[-<<+]-<<<[<]]<]>[<<<++++++>>>[>]>>>>+[->>+]<<-<<+[-<<+]-<<<[<]]<]>[<<<+++++>>>[>]>>>>+[->>+]+>>-<<[-<<+]-<<<[<]]<]>[<<<++++>>>[>]>>>>+[->>+]->-<<<+[-<<+]-<<<[<]]<]>[<<<+++>>>[>]>>>>+[->>+]->+<<<+[-<<+]-<<<[<]]<]>[<++[>]>>>>+[->>+]->[<<<+[-<<+]-<<<[<]-[<<-[>->-[<+]]<+[->>[<]]<-[>-->+[<++]]<++[-->>[<]]<++>>[[-<+>]<<[->>+<<]]<[>]>]]<[<<+[-<<+]-<<<[<]>--<<++>]>]<]>[<<<+>>>[>]>>>>+[->>+]->[<<<+[-<<+]-<<<[<]]<[<<+[-<<+]-<<<[<]+[>-[<-<]<<[>>]>>-[<+<]<<[>>]>>++<[>[-<<+>>]<[->+<]]<[>]>]]>[[-<<+>>]<[->+<]>]]>]
\$\endgroup\$
0
11
+500
\$\begingroup\$

Vyxal, 150 149 148 bytes

k2T0ẋ→_0→0?£{D¥L<|¥i:‛+-$c[:‛ +ḟ←_:_←›Ǔṫ∇∇+J←›ǔ→_|:‛<>$c[:‛ >ḟ←+→|:\.=[←_← iC₴|:\,=[←_:_←?CȦ→_|:\[=[←_← i¬[Ȯ¥$ȯ\]ḟ∇∇+$]|:\]=[Ȯ¥$Ẏf\[=TG‹∇$_]]]]]]]_›

Try it Online!

"But there's already a Vyxal answer that's 100 99 98 bytes shorter bro what is this cringe" I hear you say. Well this version doesn't use eval, and instead runs it manually. Note that it can be slow because it's doing things like rotating a 30000 item list quite frequently, so here's a version with only 100 cells for testing

Some assumptions this program makes:

  • Closed brackets
  • EOF handled manually
  • Program on first line, each input character on a new line

Explained

Quick Overview

k2T0ẋ→_                                                                                                                                                      # Tape
       0→                                                                                                                                                    # Cell pointer
         0                                                                                                                                                   # Instruction pointer
          ?£                                                                                                                                                 # Prog
            {D¥L<|                                                                                                                                           # While the instruction pointer is less than program length
                  :,←_,¥i                                                                                                                                    # Get command
                         :‛+-$c[:‛ +ḟ←_:_←›Ǔṫ∇∇+J←›ǔ→_|                                                                                                      # Handle addition and subtraction in the same place
                                                       :‛<>$c[:‛ >ḟ←+→|                                                                                      # Handle moving the pointer left and right
                                                                       :\.=[←_← iC,|                                                                         # Output
                                                                                    :\,=[←_:_←?CȦ→_|                                                         # Input
                                                                                                    :\[=[                                                    # jump to next `]` if tape[cell] == 0
                                                                                                         ←_← i¬[Ȯ¥$ȯ:,\]ḟ›∇+$]|
                                                                                                                               :\]=[Ȯ¥$Ẏf\[=TG›∇$_]          # Jump back to the previous `[`
                                                                                                                                                   ]]]]]]
                                                                                                                                                         _   # Remove the char from the stack
                                                                                                                                                          ›  # Next command
                                                                                                                                                           }

Detailed Explanation

k2T0ẋ→_

This is the tape. It consists of 30000 0s in a list. It's stored in a global variable called _.

0→

This is the cell pointer. It tracks which cell is being pointed to. It's stored in the ghost variable. The ghost variable is another name for the variable with no name (as in, its name is literally "" - the empty string.)

0

This is the instruction pointer. It tracks which character of the program is being executed. It's stored on the stack as the bottom of the stack. Throughout this answer, the stack is [instruction pointer, current character, ...] where ... is whatever processing is happening in each command.

This gets the program to execute and stores it in the register.

{D¥L<|

This is the main program execution loop. It calls its code while the instruction pointer is less than the length of the program. Before performing the comparison, the instruction pointer is triplicated (i.e. three copies of it are pushed to the stack). This is so that there is a copy for the comparison, for getting the current character in the program and for maintaining the value of the pointer.

¥i

This gets the character at the index of the instruction pointer and puts it on the stack. The stack is now [instruction pointer, command]. We now move on to handling the commands

Addition and Subtraction

:‛+-$c[:‛ +ḟ←_:_←›Ǔṫ∇∇+J←›ǔ→_

The above snippet is the entirety of the section that handles the + and - commands.

:‛+-$c

This checks if the command is in the string "+-", while leaving a copy of the command on the stack for further comparison if needed.

[:‛ +ḟ

If the command is one of + or -, then the command is duplicated yet again, and its index in the string " +" is returned. This returns 1 for + and -1 for -, as -1 is returned for characters not in the string. The 1 or -1 acts as an offset for the current cell, and saves having to check for + and - individually. It also means + can be used for both commands instead of for addition and for subtraction.

←_:_←›Ǔ

This pushes the tape (stored in the global variable called _), the value of the cell pointer + 1 (stored in the ghost variable and then incremented) and then rotates the tape left that many times. The :_ after ←_ is needed because there seems to be a bug with list mutability when rotating. (TODO: Fix)

After the rotation, the cell that is to be incremented or decremented is at the tail of the tape.

ṫ∇∇+J

This separates the tail and the rest of the list - first it pushes tape[:-1] and then it pushes tape[-1]. It then rotates the top three items on the stack so that the order is [tape[:-1], tape[-1], offset]. The offset is then added to the tail, and the tail is then appended back to the rest of the tape.

←›ǔ→_

The tape is then rotated cell pointer + 1 times to the right to "undo" the left rotation and then placed back into the global variable called _.

Moving the Cell Pointer

|:‛<>$c[:‛ >ḟ←+→

The above snippet is the entirety of the section that handles the < and > commands.

|:‛<>$c

Just like with the + and - commands, a check is done to see if the command is in the string "<>".

[:‛ >ḟ

And also just like with the + and - commands, if the command is one of < or >, then the command is duplicated yet again, and its index in the string " >" is returned. This returns 1 for > and -1 for <. The 1 or -1 acts as an offset for the current cell location, and saves having to check for < and > individually. It also means + can be used for both commands instead of for moving right and for moving left.

←+→

This adds the offset to the cell pointer and updates the value stored in the ghost variable. It also acts as a weird face.

Output

|:\.=[←_← iC₴

The above snippet is the entirety of the section that handles the . command.

|:\.=

This checks if the command is equal to the string ".". Unlike addition/subtraction and cell pointer movement, input and output cannot be handled in the same if-statement, as their functions are different at a fundamental level.

[←_← i

If the command is ., the tape is pushed, as well as the cell pointer's value. The item at the location of the cell pointer is then retrieved. The space after the second is needed to avoid it being interpreted as ←i

C₴

This prints that item after converting it to its ASCII equivalent (think chr in python)

Input

|:\,=[←_:_←?CȦ→_

The above snippet is the entirety of the section that handles the , command.

|:\,=

This checks if the command is equal to the string ",".

[←_:_←?C

If the command is ,, the tape and current cell pointer are pushed to the stack, as well as the ordinal value (think ord in python) of the next input.

Ȧ→_

This sets the cell pointerth item of tape to the input. Basically tape[cell_pointer] = ord(input()).

Looping

|:\[=[←_← i0=[Ȯ¥$ȯ\]ḟ∇∇+$]

The above snippet is the entirety of the section that handles the [ command.

|:\[=

The usual check for a certain character. [ this time.

[←_← i

If the command is [, get the cell pointerth item of tape.

¬[

If the cell is not a truthy value (i.e. not 0), then:

Ȯ¥$ȯ\]ḟ

Push program[instruction_pointer:] and find the first ] in that string. The stack is now [instruction pointer, command, position of "]"].

∇∇+$

Rotate the stack so that its order is [command, position of "]", instruction pointer] and add the position of the ] to the instruction pointer. This has the effect of iterating through the program until the next ] is found without having to have lengthy while loops.

|:\]=[Ȯ¥$Ẏf\[=TG‹∇$_

The above snippet is the entirety of the section that handles the ] command.

|:\]=

Check if the command is ]

[Ȯ¥$Ẏf\[=TG

And if it is, get the greatest index of all [s in the string program[0:instruction_pointer]. This has the effect of backtracking to the matching [ without having to have a lengthy while loop.

‹∇$_

Decrement that so that the instruction pointer will be re-incremented to the position of the matching [ and rotate the stack so that the stack order is once again [instruction pointer, command]

Final bits

]]]]]]] # Close all the if statements
_       # Remove the command from the stack
›       # Move the instruction pointer forward 1
}       # Close the main while loop
\$\endgroup\$
1
  • 2
    \$\begingroup\$ ←_← 10/10 emoticon \$\endgroup\$
    – emanresu A
    Jun 13, 2022 at 7:59
10
\$\begingroup\$

CJam, 75 bytes

lq3e4Vc*@{"-<[],.+>"#"T1$T=(t T(:T; { _T=}g \0+(@T@t _T=o "_'(')er+S/=}%s~@

Try it online: string reverser, Hello World.

Explanation

Takes code on the first line of STDIN and input on all lines below it.

l            Read a line from STDIN (the program) and push it.
 q           Read the rest of STDIN (the input) and push it.
  3e4Vc*     Push a list of 30000 '\0' characters.
        @    Rotate the stack so the program is on top.

{               }%   Apply this to each character in prog:
 "-<[],.+>"#         Map '-' to 0, '<' to 1, ... and everything else to -1.
            ...=     Push a magical list and index from it.

s~       Concatenate the results and evaluate the resulting string as CJam code.
  @      Rotate the top three elements again -- but there are only two, so the
         program terminates.

What about that magical list?

"T1$T=(t T(:T; { _T=}g \0+(@T@t _T=o "  Space-separated CJam snippets.
                                        (Note the final space! We want an empty
                                        string at the end of the list.)
_'(')er+                                Duplicate, change (s to )s, append.
        S/                              Split over spaces.

The resulting list is as follows:

T1$T=(t    (-)
T(:T;      (<)
{          ([)
_T=}g      (])
\0+(@T@t   (,)
_T=o       (.)
T1$T=)t    (+)
T):T;      (>)
{          (unused)
_T=}g      (unused)
\0+(@T@t   (unused)
_T=o       (unused)
           (all other characters)

We generate the snippets for + and > from those for - and <, simply by changing left parens (CJam’s “decrement”) into right parens (CJam’s “increment”).

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Shortest answer & biggest winner \$\endgroup\$
    – Jack G
    May 15, 2018 at 0:32
9
\$\begingroup\$

C 284 362 (From a file)

#include <stdio.h>
char b[30000],z[9999],*p=b,c,*a,i;f(char*r,int s){while(c=*a++){if(!s){(c-62)?(c-60)?(c-43)?(c-45)?(c-46)?(c-44)?0:(*p=getchar()):putchar(*p):--*p:++*p:--p:++p;if(c==91)f(a,!*p);else if(c==93){if(!*p)return;else a=r;}}else{if(c==93){--s;if(!*p&&!s)return;}else if(c==91){s++;}}}}main(int c,char**v){fread(z,1,9999,fopen(*++v,"r"));a=z;f(0,0);}

Primes:

Primes up to: 100
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97
Press any key to continue . . .

Compiled and ran successfully VS2008

Original solution failed to recognize loops that were initially set to zero. Still some room to golf. But finally solves the Prime Number program.

Ungolfed:

#include <stdio.h>
char b[30000],z[9999],*p=b,c,*a,i;
f(char*r,int s)
{
    while(c=*a++)
    {   
        if(!s)
        {
            (c-62)?(c-60)?(c-43)?(c-45)?(c-46)?(c-44)?0:(*p=getchar()):putchar(*p):--*p:++*p:--p:++p;
            if(c==91)f(a,!*p);
            else if(c==93){if(!*p)return;else a=r;}
        }
        else
        {
            if(c==93)
            {
                --s;
                if(!*p&&!s)return;
            }
            else if(c==91)
            {
                s++;
            }
        }
    }
}

main(int c,char**v){
    fread(z,1,9999,fopen(*++v,"r"));
    a=z;
    f(0,0);
}

Tests:

Hello World

Rot13

\$\endgroup\$
16
  • \$\begingroup\$ Are you checking the same pointer (l) every time you loop? I think you are supposed to check the current location of the head (p). \$\endgroup\$
    – Alexandru
    Jan 30, 2011 at 19:51
  • \$\begingroup\$ I pass the pointer to the buffer and the pointer to the stream. It checks at the end of the loop to see if the pointer l in the buffer has reached zero and breaks else it resets the stream back to the original loop [. This is necessary for nested [ loops. \$\endgroup\$
    – snmcdonald
    Jan 30, 2011 at 20:00
  • 1
    \$\begingroup\$ Yeah. I thought so. You should not check the value at pointer at first enter in the loop, but the value at the current pointer. Check the test in the question. Your program hangs. \$\endgroup\$
    – Alexandru
    Jan 30, 2011 at 20:10
  • 1
    \$\begingroup\$ You can replace break;else by return;. \$\endgroup\$
    – Alexandru
    Jan 30, 2011 at 20:56
  • 3
    \$\begingroup\$ I think you can replace (c==62)?a:b with (c-62)?b:a. \$\endgroup\$
    – Alexandru
    Jan 30, 2011 at 22:01
9
\$\begingroup\$

PHP 5.4, 296 294 273 263 261 209 191 183 178 166 characters:

I gave it a shot without using eval, but I eventually had to use it

<?$b=0;eval(strtr(`cat $argv[1]`,["]"=>'}',"["=>'while($$b){',"."=>'echo chr($$b);',","=>'$$b=fgetc(STDIN);',"+"=>'$$b++;',"-"=>'$$b--;',">"=>'$b++;',"<"=>'$b--;']));

All commands are working. This heavily abuses variable variables, and spews warnings. However, if one changes their php.ini to squelch warnings (or pipes stderr to /dev/null), this works great.

Verification (It's the "Hello World!" example from Wikipedia): http://codepad.viper-7.com/O9lYjl

Ungolfed, 367 365 335 296 267 characters:

<?php
$a[] = $b = 0;
$p = implode("",file($argv[1])); // Shorter than file_get_contents by one char
$m = array("]" => '}', "[" => 'while($a[$b]){',"." => 'echo chr($a[$b]);', "," => '$a[$b]=fgetc(STDIN);', "+" => '$a[$b]++;', "-" => '$a[$b]--;', ">" => '$b++;', "<" => '$b--;');
$p = strtr($p,$m);
@eval($p);

This should be run via the command line: php bf.php hello.bf

\$\endgroup\$
0
8
\$\begingroup\$

Windows PowerShell, 204

'$c=,0*3e4;'+@{62='$i++
';60='$i--
';43='$c[$i]++
';45='$c[$i]--
';44='$c[$i]=+[console]::ReadKey().keychar
';46='write-host -n([char]$c[$i])
';91='for(;$c[$i]){';93='}'}[[int[]][char[]]"$(gc $args)"]|iex

Fairly straightforward conversion of the instructions and then Invoke-Expression.

History:

  • 2011-02-13 22:24 (220) First attempt.
  • 2011-02-13 22:25 (218) 3e4 is shorter than 30000.
  • 2011-02-13 22:28 (216) Unnecessary line breaks. Matching on integers instead of characters is shorter.
  • 2011-02-13 22:34 (207) Used indexes into a hash table instead of the switch.
  • 2011-02-13 22:40 (205) Better cast to string removes two parentheses.
  • 2011-02-13 22:42 (204) No need for a space after the argument to Write-Host.
\$\endgroup\$
8
\$\begingroup\$

F#: 489 chars

The following program doesn't jump at '[' / ']' instructions, but scans the source code for the next matching token. This of course makes it kind of slow, but it can still find the primes under 100. F# integer types don't overflow but wrap.

Here's the short version:

[<EntryPoint>]
let M a=
 let A,B,i,p,w=Array.create 30000 0uy,[|yield!System.IO.File.ReadAllText a.[0]|],ref 0,ref 0,char>>printf"%c"
 let rec g n c f a b=if c then f i;if B.[!i]=a then g(n+1)c f a b elif B.[!i]=b then(if n>0 then g(n-1)c f a b)else g n c f a b
 while !i<B.Length do(let x=A.[!p]in match B.[!i]with|'>'->incr p|'<'->decr p|'+'->A.[!p]<-x+1uy|'-'->A.[!p]<-x-1uy|'.'->w x|','->A.[!p]<-byte<|stdin.Read()|'['->g 0(x=0uy)incr '['']'|']'->g 0(x>0uy)decr ']''['|_->());incr i
 0

A nasty gotcha was that the primes.bf program chokes on windows newlines. In order to run it I had to save the input number to a UNIX formatted text document and feed it to the program with a pipe:

interpret.exe prime.bf < number.txt

Edit: entering Alt+010 followed by Enter also works in Windows cmd.exe

Here's the longer version:

[<EntryPoint>]
let Main args =
    let memory = Array.create 30000 0uy
    let source = [| yield! System.IO.File.ReadAllText args.[0] |]
    let memoryPointer = ref 0
    let sourcePointer = ref 0
    let outputByte b = printf "%c" (char b)
    let rec scan numBraces mustScan adjustFunc pushToken popToken =
        if mustScan then
            adjustFunc sourcePointer
            if source.[!sourcePointer] = pushToken then
                scan (numBraces + 1) mustScan adjustFunc pushToken popToken
            elif source.[!sourcePointer] = popToken then
                if numBraces > 0 then scan (numBraces - 1) mustScan adjustFunc pushToken popToken
            else
                scan numBraces mustScan adjustFunc pushToken popToken 

    while !sourcePointer < source.Length do
        let currentValue = memory.[!memoryPointer]
        match source.[!sourcePointer] with
            | '>' -> incr memoryPointer
            | '<' -> decr memoryPointer
            | '+' -> memory.[!memoryPointer] <- currentValue + 1uy
            | '-' -> memory.[!memoryPointer] <- currentValue - 1uy
            | '.' -> outputByte currentValue
            | ',' -> memory.[!memoryPointer] <- byte <| stdin.Read()
            | '[' -> scan 0 (currentValue = 0uy) incr '[' ']'
            | ']' -> scan 0 (currentValue > 0uy) decr ']' '['
            |  _  -> ()
        incr sourcePointer
    0 
\$\endgroup\$
2
  • \$\begingroup\$ I solved the Enter issue by not pressing it but Ctrl+J :-) \$\endgroup\$
    – Joey
    Feb 14, 2011 at 8:11
  • \$\begingroup\$ Ctrl+J didn't work for me, but entering Alt+010 followed by Enter did. \$\endgroup\$
    – cfern
    Feb 14, 2011 at 9:28
8
\$\begingroup\$

C, 333 characters

This is my first BF interpreter and the first golf I actually had to debug.

This runs the prime number generator on Mac OS X/GCC, but an additional #include<string.h> may be necessary at a cost of 19 more characters if the implicit definition of strchr doesn't happen to work on another platform. Also, it assumes O_RDONLY == 0. Aside from that, leaving int out of the declaration of M saves 3 characters but that doesn't seem to be C99 compliant. Same with the third * in b().

This depends on the particulars of ASCII encoding. The Brainfuck operators are all complementary pairs separated by a distance of 2 in the ASCII code space. Each function in this program implements a pair of operators.

#include<unistd.h>
char C[30000],*c=C,o,P[9000],*p=P,*S[9999],**s=S,*O="=,-\\",*t;
m(){c+=o;}
i(){*c-=o;}
w(){o<0?*c=getchar():putchar(*c);}
b(){if(o>0)*c?p=*s:*--s;else if(*c)*++s=p;else while(*p++!=93)*p==91&&b();}
int(*M[])()={m,i,w,b};
main(int N,char**V){
read(open(V[1],0),P,9e3);
while(o=*p++)
if(t=strchr(O,++o&~2))
o-=*t+1,
M[t-O]();
}
\$\endgroup\$
3
  • \$\begingroup\$ I think you can shrink it more by using the 'e' notation for all the big numbers. \$\endgroup\$ Aug 16, 2011 at 4:34
  • \$\begingroup\$ @luser: I was initially surprised too, but the language and compiler won't allow that. I did manage to shrink another 4 chars with tweaks, and using a #define instead of the function table would also probably be terser. I just like the number 333 and the table :v) . \$\endgroup\$ Aug 16, 2011 at 5:56
  • \$\begingroup\$ Oh, right. I really should've known that. E-notation is in the production for a floating-point constant, whereas a declaration requires an integer. BTW, this may be cheating, but check out nieko.net/projects/brainfuck for Urban Müller's version. The biggest gain appears to be heavy use of ||. \$\endgroup\$ Aug 16, 2011 at 6:10
8
\$\begingroup\$

C, 267

#define J break;case
char*p,a[40000],*q=a;w(n){for(;*q-93;q++){if(n)switch(*q){J'>':++p;J'<':--p;J'+':++*p;J'-':--*p;J'.':putchar(*p);J',':*p=getchar();}if(*q==91){char*r=*p&&n?q-1:0;q++;w(r);q=r?r:q;}}}main(int n,char**v){p=a+read(open(v[1],0),a,9999);*p++=93;w(1);}

Run as ./a.out primes.bf

Ungolfed Version:

#define J break;case

char*p,a[40000],*q=a; // packed so program immediately followed by data

w(n){
    for(;*q-93;q++){ // until ']'
        if(n)switch(*q){ // n = flagged whether loop evaluate or skip(0)
                J'>':++p;
                J'<':--p;
                J'+':++*p;
                J'-':--*p;
                J'.':putchar(*p);
                J',':*p=getchar();
        }
        if(*q==91){char*r=*p&&n?q-1:0;q++;w(r);q=r?r:q;} // recurse on '[', record loop start
    }
}

main(int n,char**v){
    p=a+read(open(v[1],0),a,9999);
    *p++=93; // mark EOF with extra ']' and set data pointer to next
    w(1); // begin as a loop evaluate
}
\$\endgroup\$
8
\$\begingroup\$

C, 260 + 23 = 283 bytes

I created a C program which can be found here.

main(int a,char*s[]){int b[atoi(s[2])],*z=b,p;char*c=s[1],v,w;while(p=1,
*c){q('>',++z)q('<',--z)q('+',++*z)q('-',--*z)q('.',putchar(*z))q(',',*z
=getchar())if(*c=='['||*c==']'){v=*c,w=184-v;if(v<w?*z==0:*z!=0)while(p)
v<w?c++:c--,p+=*c==v?1:*c==w?-1:0;}c++;}}

Has to be compiled via gcc -D"q(a,b)"="*c-a||(b);" -o pmmbf pmmbf.c and can be called as follows: pmmbf ",[.-]" 30000 whereby the first argument (quoted) contains the bf-program to run, the second determines how large the tape should be.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ I think that you need to add 23 characters to your count for the -D"q(a,b)"="*c-a||(b);" option, since that seems (to my limited understanding, at least) to be helping you shrink your code. \$\endgroup\$
    – Gareth
    Aug 5, 2011 at 9:23
  • \$\begingroup\$ The option is included in the posted text. The reason for it is to avoid the lengthy word define and newline, but I don't think that's really kosher. Anyway with the quotes, comment, and gcc -D I don't see the advantage at all. \$\endgroup\$ Aug 6, 2011 at 3:30
7
\$\begingroup\$

Delphi, 397 382 378 371 366 364 328 characters

Eat this Delphi!

328 var p,d:PByte;f:File;z:Word=30000;x:Int8;begin p:=AllocMem(z+z);d:=p+z;Assign(F,ParamStr(1));Reset(F,1);BlockRead(F,p^,z);repeat z:=1;x:=p^;case x-43of 1:Read(PChar(d)^);3:Write(Char(d^));0,2:d^:=d^+44-x;17,19:d:=d+x-61;48,50:if(d^=0)=(x=91)then repeat p:=p+92-x;z:=z+Ord(p^=x)-Ord(p^=x xor 6);until z=0;end;Inc(p)until x=0;end.

Here the same code, indented and commented :

var
  d,p:PByte;
  x:Int8;
  f:File;
  z:Word=30000;
begin
  // Allocate 30000 bytes for the program and the same amount for the data :
  p:=AllocMem(z+z);
  d:=p+z;
  // Read the file (which path must be specified on the command line) :
  Assign(F,ParamStr(1));
  Reset(F,1);
  BlockRead(F,p^,z);
  // Handle all input, terminating at #0 (better than the spec requires) :
  repeat
    // Prevent a begin+end block by preparing beforehand (values are only usable in '[' and ']' cases) :
    z:=1;                       // Start stack at 1
    x:=p^;                      // Starting at '[' or ']'
    // Choose a handler for this token (the offset saves 1 character in later use) :
    case x-43of
      1:Read(PChar(d)^);        // ','     : Read 1 character from input into data-pointer
      3:Write(Char(d^));        // '.'     : Write 1 character from data-pointer to output
      0,2:d^:=d^+44-x;          // '+','-' : Increase or decrease data
      17,19:d:=d+x-61;          // '<','>' : Increase or decrease data pointer
      48,50:                    // '[',']' : Start or end program block, the most complex part :
        if(d^=0)=(x=91)then     // When (data = 0 and forward), or when (data <> 0 and backward)
        repeat                  //
          p:=p+92-x;            // Step program 1 byte back or forward
          z:=z+Ord(p^=x)        // Increase stack counter when at another bracket
              -Ord(p^=x xor 6); // Decrease stack counter when at the mirror char
        until z=0;              // Stop when stack reaches 0
    end;
    Inc(p)
  until x=0;
end.

This one took me a few hours, as it's not the kind of code I normally write, but enjoy!

Note : The prime test works, but doesn't stop at 100, because it reads #13 (CR) before #10 (LF)... do other submissions suffer this problem too when running on CRLF OSes?

\$\endgroup\$
1
  • \$\begingroup\$ Wow! I never would have expected to trump C in terseness with Delphi! Not until you apply my ideas to C I guess ;-) \$\endgroup\$
    – PatrickvL
    Mar 12, 2011 at 16:30
7
\$\begingroup\$

Python 2, 223

I admit that I recycled an old program of mine (but had to change it quite a bit, because the old version didn't have input, but error checking...).

P="";i,a=0,[0]*30000
import os,sys
for c in open(sys.argv[1]).read():x="><+-.[,]".find(c);P+=" "*i+"i+=1 i-=1 a[i]+=1 a[i]-=1 os.write(1,chr(a[i])) while+a[i]: a[i]=ord(os.read(0,1)) 0".split()[x]+"\n";i+=(x>4)*(6-x)
exec P

Runs the primes calculator fine.

I see now that Alexandru has an answer that has some similarities. I'll post mny answer anyways, because I think there are some new ideas in it.

\$\endgroup\$
7
\$\begingroup\$

Desmos, 470 bytes

Program input:

  • variable \$P\$, containing an array of ASCII codepoints representing the brainfuck program.
  • variable \$I\$, also containing an array of ASCII codes for stdin

You can use this program to convert your ASCII to a list of numbers. Unless you like to play the waiting game, I would also recommend that you minify your program

stdout is represented by the array \$O\$. Like \$P\$ and \$I\$, it is also an array of codepoints.

In addition, you can inspect the memory tape in variable \$M\$. The memory consists of 30000 unsigned bytes, though it will only store as many bytes as necessary because storing 30000 bytes is not great for performance.

To begin execution, run the "reset" action and then start the ticker.


Ticker code

\left\{j<=P.length:c,a\right\}

Expression list

O=[]
i=1
k=1
M=[0]
s(x)=M->[\{l=p:\mod(x,256),M[l]\}\for l=[1...M.\length]]
p=1
v=M[p]
w=P[i]
j=1
N=[]
S=[]
a=\{59<w<63:p->p+w-61,w=44:b,42<w<46:s(v+44-w),w=46:O->O.\join(v),w=91:i->\{v=0:N[i],i\}+1,w=93:i->\{v=0:i,N[i]\}+1\},\{w=62:\{30000>p>=M.\length:M->M.\join(0)\}\},\{\{w=91,w=93,0\}=0:i->i+1\}
b=s(I[k]),k->k+1
c=\{P[j]=93:d,N->N.\join(0)\},\{P[j]=91:S->\join(j,S)\},j->j+1
d=N->[\{m=S[1]:j,m=j:S[1],N[m]\}\for m=[1...j]],S->S[2...]

Try it on Desmos!

The program loaded in the above TioD is a "Hello, World!" program. The prime number program took around five minutes to print 2 3 5 so I decided that maybe it wasn't the best idea to run it.


How it works

The interpreter first matches up bracket pairs, then scans through the program character-by-character and executes them.

Below is the code for bracket-matching:

j=1
N=[]
S=[]
c=\{P[j]=93:d,N->N.\join(0)\},\{P[j]=91:S->\join(j,S)\},j->j+1
d=N->[\{m=S[1]:j,m=j:S[1],N[m]\}\for m=[1...j]],S->S[2...]
  • \$j\$ is a list index, indicating that character \$P[j]\$ is currently being processed
  • \$N\$ holds the index of the matching bracket pair, if it exists. If \$N[i]\$ is a positive integer, then \$P[i]\$ matches with \$P[N[i]]\$.
  • \$S\$ is a stack containing the indices of all unmatched [s thus far. The top of the stack is at \$S[1]\$.

Next there are two actions, \$c\$ and \$d\$.

Every time action \$c\$ is run, the next character in the program is processed.

c=\{P[j]=93:d,N->N.\join(0)\},\{P[j]=91:S->\join(j,S)\},j->j+1
c=                                                                c is an action containing:
  \{                       \},                                     a conditional...
    P[j]=93:d,                                                      that runs d if P[j]=93 (']')
              N->N.\join(0)                                         and appends 0 to N otherwise
                              \{                     \},           a conditional...
                                P[j]=91:                            that when P[j]=91 ('[')...
                                        S->\join(j,S)                pushes j to S
                                                        j->j+1     increment j

\$d\$ is the response to when a ']' character is detected. Because it consists of two actions, it cannot be directly placed in the conditional and thus must be defined separately.

d=N->[\{m=S[1]:j,m=j:S[1],N[m]\}\for m=[1...j]],S->S[2...]
d=                                                            d is an action consisting of:
  N->[\{                  N[m]\}\for m=[1...j]],               a modification to N
        m=S[1]:j,                                               in which N[S[1]] is set to j
                 m=j:S[1],                                      and N[j] is set to S[1]
                                                S->S[2...]     the topmost value is popped from S

Once bracket pairs are matched the expression is evaluated character-by-character:

i=1
k=1
M=[0]
s(x)=M->[\{l=p:\mod(x,256),M[l]\}\for l=[1...M.\length]]
p=1
v=M[p]
w=P[i]
a=\{59<w<63:p->p+w-61,w=44:b,42<w<46:s(v+44-w),w=46:O->O.\join(v),w=91:i->\{v=0:N[i],i\}+1,w=93:i->\{v=0:i,N[i]\}+1\},\{w=62:\{p=M.\length:M->M.\join(0)\}\},\{\{w=91,w=93,0\}=0:i->i+1\}
b=s(I[k]),k->k+1
  • \$i\$ is the instruction pointer
  • \$k\$ is the next character to be read in stdout
  • \$M\$ is the memory tape
  • \$s(x)\$ sets the value of the current cell to \$x\$
  • \$p\$ points to the current memory cell
  • \$v\$ is the value of the current cell
  • \$w\$ is the current instruction

\$a\$ is an action that executes the next character. \$b\$ is a utility for reading from stdin - like \$d\$, \$b\$ is separated because conditional branches only support one action.

\$a\$ is rather long, so here it is branch-by-branch:

a=
\{
  59<w<63:p->p+w-61,                       ><: hack for p±±-ing; 6 bytes smaller than doing it separately
  w=44:b,                                  ,: read from stdin (do this before +- to avoid interference)
  42<w<46:s(v+44-w),                       +-: same hack
  w=46:O->O.\join(v),                      .: write to stdout
  w=91:i->\{v=0:N[i],i\}+1,                [: jump to end if v=0
  w=93:i->\{v=0:i,N[i]\}+1                 ]: jump to start if v!=0
\},
\{w=62:\{p=M.\length:M->M.\join(0)\}\},    widen memory tape if necessary
\{\{w=91,w=93,0\}=0:i->i+1\}               increment instruction pointer if not '[' or ']'

Essentially \$a\$ is a big conditional that performs different actions for different characters, then ties up some loose ends afterwards.

\$b\$, which reads from stdin, is relatively straightforward:

b=s(I[k]),k->k+1
b=                  b is an action that
  s(I[k]),          writes the next character from stdin into memory
          k->k+1    and increments the character pointer for stdin

Finally, the exection of all of this is controlled by a ticker:

\left\{j<=P.length:c,a\right\}

This ticker will choose between \$a\$ and \$c\$ depending on whether the bracket matching is done or not.

When the input program and stdin are set up and the ticker is run, this program should be able to interpret any brainfuck expression. Just maybe not in a reasonable time frame.

\$\endgroup\$
7
\$\begingroup\$

X86_64/Linux Machine Code, 104 101 100 99 98 97 96 95 93 92 91 89

Usage:

$> gcc -s -static -nostartfiles -nodefaultlibs -nostdlib -Wl,-Tbss=0x7ffe0000 -DBF_IDEAL_BSS -Wl,--build-id=none bf.S -o bf
$> ./bf <input>

Notes

  • Size is measured in machine code (i.e 89 bytes of PROGBITS; .text, .data, etc...)
  • -9 bytes if reads input prog from stdin
  • -4 more bytes without BF_EXITCLEANLY (it will segfault to exit).
  • -2 more bytes with BF_BALANCED (assumes that over course of program data cell returns to start).
  • So minimum possible size 74 bytes.
  • NOTE Without the ideal bss setup (at address 0x7ffe0000 and BF_IDEAL_BSS defined as seen in compile command above) add +2 bytes.
#define BF_LBRACE   91
#define BF_RBRACE   93

#define BF_DOT  46
#define BF_COMMA    44

#define BF_PLUS 43
#define BF_MINUS    45

#define BF_LSHIFT   60
#define BF_RSHIFT   62

#define BF_READFILE
#define BF_EXITCLEANLY
    // #define BF_IDEAL_BSS
    // #define BF_BALANCED


    .global _start
    .text
_start:
    /* All incoming registers at zero.  */

    /* Large read range. This may cause errors for huge programs
       (which will fail anyways).  */
    decl    %edx
#ifdef BF_READFILE
    /* open.  */
    movb    $2, %al
    /* 3rd argument in rsp is first commandline argument.  */
    pop %rdi
    pop %rdi
    pop %rdi
    /* O_RDONLY is zero.  */
    syscall
    /* Error is okay, we will eventually just exit w.o executing.  */
    xchgl   %eax, %edi

    /* Sets ESI at 2 ** 31. This is an arbitrary address > 65536
       (minimum deref address on linux) < 2 ** 32 - PROG_SPACE
       (PROG_SPACE ~= 262144). We setup bss at 2 ** 31 - 131072 via
       linker script (-Tbss=0x7ffe0000).  */
# ifdef BF_IDEAL_BSS
    bts %edx, %esi
# else
    movl    $(G_mem + 65536), %esi
# endif
    /* bss initialized memory is zero. This is cheapest way to zero
       out eax.  */
    lodsl
#else
# ifdef BF_IDEAL_BSS
    bts %edx, %esi
# else
    movl    $(G_mem + 65536), %esi
# endif
#endif

    /* eax/edi are already zero which happens to match
       SYS_read/STDIN_FILENO.  */
    syscall


    /* Usage linux' pre-allocated stack for braces.  */

    /* Program code grows up.  */
    movl    %esi, %ebx
    /* Assuming no errors, reach stores size in eax. Note errors or
       0-byte reads are okay. The ebx is readable memory and zero-
       initialized (bss), so it its zero-length, it will just be an
       invalid op and we will hit bounds check below. If its error,
       eax is negative so ebx will be negative and likewise we will
       hit bounds check below.  */

#ifdef BF_BALANCED
    addl    %eax, %esi
    xchgl   %eax, %ebp
#else
    addl    %esi, %eax
    xchgl   %eax, %ebp
    movl    %ebp, %esi
#endif

    /* We have -1 in edx, so negative to get 1. 1 is needed in a
       variety of places.  */
    negl    %edx
run_program:
    /* Need to zero eax.  */
    movb    (%rbx), %al
    incl    %ebx

    /* %al contains the program "instruction". Its unique to one of
       8 values so just test each in order. If instruction matches
       execute it then fallthrough (%al can only ever match one).
       We occasionally set al but are sure never to set it to the
       ASCII value of any of our 8 instructions.  */

    /* TODO: Brace handling could probably be smaller.  */
try_lbrace:
    cmpb    $BF_LBRACE, %al
    je  do_lbrace
try_rbrace:
    cmpb    $BF_RBRACE, %al
    jne try_cont
do_rbrace:
    /* Popping state (we might repush if we are looping back).  */
    pop %rdx
    pop %rdi
    /* Non-zero cell means loop. Note we have 1 cached in edx.  */
    cmpb    %dl, (%rsi)
    jb  next_insn
    movl    %edi, %ebx
do_lbrace:
    /* Restore loop state.  */

    push    %rbx
    push    %rdx

    /* If cell is zero, then we are skipping till RBRACE. Note we
       have 1 cached in edx.  */
    cmpb    %dl, (%rsi)
    /* -1 if we want to start skipping.  */
    sbb %dh, %dh
try_cont:
    orb %dh, %al
    /* we have set ah s.t its either zero or has a value that makes
       any further matches impossible.  */

    /* For the rest of the ops we take advantage of the fact that
       the ascii values of the pairs '<'/'>', '+'/'-', and '.',','
       are all 2 apart. This allows use to test a pair with the
       following formula: `((al - PAIR_LO) & -3) == 0`. This will
       always leave the pair as 0/2 and will match only the pair.
       It turns out 0/2 are useful and can be used to do all the
       rest of the operations w.o extra branches.  */

try_lshift:
    subb    $BF_LSHIFT, %al
    testb   $-3, %al
    jnz try_comma
    addl    %eax, %esi
    decl    %esi
try_comma:
try_dot:
    subb    $(BF_PLUS - BF_LSHIFT), %al

    /* We have 0/1/2/3 for '+'/','/'-'/'.' so check if we remaining
       valid opcode.  */
    cmpb    $3, %al
    ja  next_insn

    /* 0/2 are '+'/'-' so check low bits. TODO: There may be a way
       to just care from `shr $1, %al' which would save us an
       instruction. But it messes up the '+'/'-' case.  */
    testb   $-3, %al
    jz  try_minus

    /* al is either 3/1. Shift by 1 to get 1/0 for our syscall number.  */
    shrb    $1, %al

    // btsq    $, %rax
    /* SYS_write == STDOUT_FILENO, SYS_read == STDIN_FILENO.  */
    movl    %eax, %edi
    /* We already have 1 in rdx.  */
    syscall
    /* Assuming no io error, eax is 1. We will subtract 1 from eax
       in try_minus/try_plus so it will be +/- 0 which does
       nothing.  */
try_minus:
try_plus:
    /* If not coming from syscall eax is 0/2. 0 --> plus, 2 -->
       minus. So (rsi - eax + 1) translates.  */
    // setbe   %cl

    subb    %al, (%rsi)
    incb    (%rsi)

next_insn:
#ifdef BF_BALANCED
    /* If BF_BALANCED is set, we assume that the program will have
       shifted back columns to start before exiting.  */
    cmpl    %ebx, %esi
#else
    cmpl    %ebx, %ebp
#endif
    jg  run_program

#ifdef BF_EXITCLEANLY
    /* eax has zero upper 24 bits so we can cheat and use movb here.
       (This isn't exact correct, assuming no IO errors on last
       instruction).  */
    movb    $60, %al
    syscall
#endif

    .section .bss
    .align  32
G_mem:  .space(65536 * 4)

Edit: Just for fun, here a compliant 269 byte fully contained ELF file that implements brainfuck on Linux/X86_64:

Its possible to make this smaller.

  • We could take advantage of the known linux ELF impl and put code in known-unused fields.
00000000  7f 45 4c 46 02 01 01 00  00 00 00 00 00 00 00 00  |.ELF............|
00000010  02 00 3e 00 01 00 00 00  b0 10 40 00 00 00 00 00  |..>.......@.....|
00000020  40 00 00 00 00 00 00 00  00 00 00 00 00 00 00 00  |@...............|
00000030  00 00 00 00 40 00 38 00  02 00 00 00 00 00 00 00  |[email protected].........|
00000040  01 00 00 00 05 00 00 00  00 00 00 00 00 00 00 00  |................|
00000050  00 10 40 00 00 00 00 00  00 10 40 00 00 00 00 00  |..@.......@.....|
00000060  cd 00 00 00 00 00 00 00  cd 00 00 00 00 00 00 00  |................|
00000070  00 00 00 00 00 00 00 00  01 00 00 00 06 00 00 00  |................|
00000080  00 00 00 00 00 00 00 00  00 00 fe 7f 00 00 00 00  |................|
00000090  00 00 fe 7f 00 00 00 00  00 00 00 00 00 00 00 00  |................|
000000a0  00 00 04 00 00 00 00 00  20 00 00 00 00 00 00 00  |........ .......|
000000b0  ff ca b0 02 5f 5f 5f 0f  05 97 0f ab d6 ad 0f 05  |....___.........|
000000c0  89 f4 89 f3 01 c6 89 f5  f7 da 0f b6 03 ff c3 3c  |...............<|
000000d0  5b 74 0c 3c 5d 75 0e 5a  5f 38 16 72 28 89 fb 53  |[t.<]u.Z_8.r(..S|
000000e0  52 38 16 18 f6 08 f0 2c  3c a8 fd 75 04 01 c6 ff  |R8.....,<..u....|
000000f0  ce 2c ef 3c 03 77 0e a8  fd 74 06 d0 e8 89 c7 0f  |.,.<.w...t......|
00000100  05 28 06 fe 06 39 dd 7f  c1 b0 3c 0f 05           |.(...9....<..|
0000010d
\$\endgroup\$
19
  • 1
    \$\begingroup\$ If you're assuming 32-bit addresses, might as well link as a non-PIE executable so you can use mov $G_mem+65536, %esi (5 bytes) instead of 6 bytes for an LEA with 32-bit operand-size. (How to load address of function or label into register for gcc/clang -fno-pie code-gen). Also, at least one other asm answer reads input from stdin so I think the 93 byte size should be in a header as well. \$\endgroup\$ Aug 15, 2023 at 21:15
  • \$\begingroup\$ Do you need to exit cleanly? You could just fall off the end and segfault (which will happen reliably since EAX is a small number, so 00 00 add %al, (%rax) will segfault). \$\endgroup\$ Aug 15, 2023 at 21:19
  • \$\begingroup\$ Also, can we use stack space instead of .bss? Or no, sub $65536, %rsp / push %rsp / pop %rsi is 5 + 1 +1 = 7 bytes, and all address-copying after that will have to use 64-bit operand-size since the stack is outside the low 32, unless using the x32 ABI. It feels a bit like cheating to use BSS space without accounting any bytes for the extra ELF metadata to reserve it, esp. for a whole program, rather than a function where we can justify incremental size as part of a larger program. Also, are you aligning your .bss space so the low 16 bits of the address is zero? I'm still reading... \$\endgroup\$ Aug 15, 2023 at 21:22
  • 1
    \$\begingroup\$ @PeterCordes "Also, can we use stack space instead of .bss? Or no, sub $65536, %rsp / push %rsp / pop %rsi is 5 + 1 +1 = 7"(2) I think it increases code size. Particularly just in setup we need 2x leaq which is enough to make it break even. With the 64-bit inc/dec its a lost cause. \$\endgroup\$
    – Noah
    Aug 15, 2023 at 21:46
  • 1
    \$\begingroup\$ Figured it out. Since edx comes in as zero, decl %edx to read file (can overread, but if file was too large for buffer would be buggy anyways), then just negl %edx to get 1. Saves the byte :) \$\endgroup\$
    – Noah
    Sep 19, 2023 at 18:45
6
\$\begingroup\$

Recall, 594 bytes

In short: Recall has no arithmetic operators in a classic sense, it only has bitwise operations. You can not just "add one" etc. Recall is also strictly stack-based.

DC505M22022M32032M606M42042M707M92092M4405022o032o06o042o07o092o044o1305022o06o042o092o52052q.q2305022o06o07o93093q.q5403206o07o14014q.q6403206o042o07o24024q.q74Yx34034z03MMMMMMMM034o3yY030401r3.4.101zyY040301r4.3.101zY01052gZ02Z040301052023s4.3.10zyY01023gZ02z030401023052s3.4.10zyY01093gZ02q20zyY01054gZ02u20zyY01014gZx20zyY01064gZ02X0zyY01024gZ03304302r33.43.20zyY01074gZ04303302r43.33.20zyyQ6205.8Y06208g6206208iZ08M808013izy062U7205.9Y07209g7207209iz09M909013izy072R53.63.82063MMMMMMMM053o63082013i53082KKKKKKKK82053063082S84.94.12.73.83t012073083TY083073012r83.73.12012084gzY012094gZt0zyy

Example 1: Print something

Input:

-[--->+<]>-----..-[----->+<]>.++++.+[->++++<]>.---[----->++<]>.---.------------.++++++++.++++++++.+[-->+++++<]>-.

Output:

PPCG rocks!

Example 2: Output square numbers up to 100

Input:

+[>++<-]>[<+++++>-]+<+[>[>+>+<<-]++>>[<<+>>-]>>>[-]++>[-]+>>>+[[-]++++++>>>]<<<[[<++++++++<++>>-]+<.<[>----<-]<]<<[>>>>>[>>>[-]+++++++++<[>-<-]+++++++++>[-[<->-]+[<<<]]<[>+<-]>]<<-]<<-]

Output:

0
1
4
9
16
25
36
49
64
81
100

This example might take a few minuted to execute and might cause a "this tab is frozen" message. Ignore that and wait.

\$\endgroup\$
0
6
\$\begingroup\$

C (gcc), 273 268 bytes

main(_,a){_=fopen("w.c","w");fputs("main(){char a[30000],*p=a;",_);x:a=getchar();fputs(a-62?a-60?a-43?a-45?a-46?a-44?a-91?a-93?~a?"":"}":"}":"while(*p){":"*p=getchar();":"putchar(*p);":"--*p;":"++*p;":"--p;":"++p;",_);if(~a)goto x;fclose(_);system("cc w.c;./a.out");};

Try it online!

-5 thanks to ceilingcat

Takes input from stdin.

This relies a little bit on the environment, but is pretty consistent. This is effectively the eval solution for c. It writes an appropriate C program to the file w.c, compiles it, and runs it as the desired executable. Thus as a bonus effect this actually compiles the bf code and leaves a.out as a binary for it. Note that depending on the system you may need to modify the last string. In particular most windows c compilers call the default executable "a.exe". Luckily as far as I can tell, they all have the same length so the bytecount is the same. (though if you don't have a cc defined you may need to add a letter such as gcc to the compile command, adding 1 byte).

I am aware that this thread is a bit old, but I didn't see this style of C solution yet, so I thought I'd add it.

\$\endgroup\$
1
6
\$\begingroup\$

Vyxal DOb, 54 53 50 bytes

`><+-][.,`£¥↔¥`‟„›‹}`f`{:|:C₴_¼`²JĿ?ṘƛC⅛;k23*(0)„Ė

Try it Online!

-4 thanks to Aaron Miller.

The basic idea is that a stack is a wrapping tape if you can rotate the whole thing. So it's really just setting up the input, pushing 0s, and transpiling:

  • > - - rotate stack right
  • < - - rotate stack left
  • + - - increment the current item
  • - - - decrement the current item
  • [ - {:| - set a while loop without popping the top
  • ] - } - close a while loop
  • . - :C₴ - Duplicate, make a char and output without a trailing newline
  • , - - pop, then take a character from input

?ṘƛC⅛; takes the input and pushes each character to the global array. Then k23*(0) pushes 30k zeroes, then „Ė runs the code!

\$\endgroup\$
8
  • \$\begingroup\$ 52 bytes because why not \$\endgroup\$ Jul 15, 2021 at 12:58
  • \$\begingroup\$ @AaronMiller That doesn't seem to work - it outputs 0 instead of a. Nice idea though! \$\endgroup\$
    – emanresu A
    Jul 15, 2021 at 20:14
  • \$\begingroup\$ I swear it worked perfectly this morning. Somebody must've broken it sometime today. \$\endgroup\$ Jul 15, 2021 at 20:16
  • \$\begingroup\$ @AaronMiller Lyxal.... I'll try it on hyper's fork \$\endgroup\$
    – emanresu A
    Jul 15, 2021 at 20:31
  • 1
    \$\begingroup\$ Until it gets fixed, here's 53 bytes \$\endgroup\$ Jul 15, 2021 at 20:36

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