Print the ASCII Code page

Question

Goal

You have to print the ASCII printable characters' code page (0x20-0x7E). The output should look like this:

 !"#$%&'()*+,-./
0123456789:;<=>?
@ABCDEFGHIJKLMNO
PQRSTUVWXYZ[\]^_
`abcdefghijklmno
pqrstuvwxyz{|}~

Rules

• The output should be exactly like shown above, but trailing whitespace is OK.
• No builtins that trivialize the question (e.g. the ones that print the ASCII table) ! They can be posted for documentation purposes, but will not be in the leaderboards.
• Standard rules for I/O and loopholes apply.
• Lowest bytecount wins.

Good Luck!

P.S. This challenge is well-known, but a straight "print every single printable ASCII character" has somehow never been done here.

Answer 1

Python 3, 42 bytes

print(('%c'*16+'\n')*6%(*range(32,127),9))

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Includes a trailing tab on the last line and a trailing newline.

Answer 2

Python 3, 52 bytes

for i in range(95):print(chr(i+32),end='\n'[~i%16:])

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Answer 3

Prelude, ⁵⁰ 42 bytes

88+
6(1-) v^^(  1-++!)^6-!)
1- v+(1-) v(1-

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Explanation

88+
6(1-)
1- v+

Push \$16\$ to Voice 1 and \$95\$ to Voice 3.

       v^^
      (1-)

For each value from \$95\$ down to \$1\$, push that value and two copies of \$16\$ to Voice 2. When the loop ends, the stack in Voice 2 is [..., 0, 0, 0, 95, 16, 16, 94, 16, 16, ..., 1, 16, 16].

         (  1-++!     )

For each triplet of values in Voice 2, sum them and subtract \$1\$, then print the result as an ASCII character...

                 )
          v(1-

in groups of 16...

                  ^6-!

with a newline appended to each group.

Printing fails on the (non-existent) 96th character because the triplet of values summed is [0, 0, 0], with the result that the code attempts to convert \$-1\$ to ASCII and thus exits with an out-of-range error.

---

Fugue

Fugue is a musical encoding for Prelude. Fugascii*, a short trio for woodwinds, was generated from the Prelude code above using a script that I wrote for this purpose. Read all about it in this answer.

^{* The title is a pun on a previous Fugue composition of mine, Fugacity.}

Answer 4

Zsh, 26 bytes

printf %s {\ ..~}|fold -16

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Answer 5

Piet + ascii-piet, 87 82 bytes (3×28=84 codels)

Thanks @Bubbler for -5 bytes by changing 4 dup * 2 * to a more space efficient 8 4 *

eeeumtqcsqrrjlvuddtrjcqdjes_eeeu ?????rr?jujrnvmsjiqqc _eeuu kdvtrbbbbb???vv   qq?

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Might add a more detailed explanation later if I feel like to, but here's an explanation image generated by Bubbler's Piet interpreter for now. Can 100% be golfed, but I'm too lazy to do that right now lol.

Explanation image

The basic algorithm is to initially push 32 to the stack, then keep incrementing and outputting the character until 127, after which the code stops. At each iteration, it checks if the number+1 is a multiple of 16, and prints a newline if it is.

Longer explanation

Here's the ascii piet version of the code, with newlines added for clarity:

eeeumtqcsqrrjlvuddtrjcqdjes_
eeeu ?????rr?jujrnvmsjiqqc _
eeuu kdvtrbbbbb???vv   qq?

I'll be going over the code section by section.

Initialization:

eeeum
eeeu
eeuu

Push 8, Push 4, multiply. This puts 32 onto the stack (the codepoint for space)

Top half of the loop:

     tqcsqrrjlvuddtrjcqdje

What it does:

Commands                              Stack          DP,     CC
dup                                   [32,32]        right   left
out (char)                            [32]           right   left
1 + Dup                               [33,33]        right   left
4 dup * dup * 2 / 1 -                 [33,33,127]    right   left
- ! !                                 [33,1]         right   left
DP+                                   [33]           down    left

Basically, it adds 1 to the current codepoint and checks if it is equal to 127. If it is, then DP+ will turn the pointer 0 times and terminate the code. Otherwise, DP+ will turn the pointer 1 time and continue the loop.

Bottom right portion of the loop:

            ?jujrnvmsjiqqc
               ???vv   qq?

What it does:

Command                              Stack          DP,     CC
dup                                  [33,33]        down    left
4 dup *                              [33,33,16]     left    right
%                                    [33,1]         left    right
! 3 *                                [33,0]         left    right
1 CC+                                [33,0]         left    left
DP+                                  [33]           left    right

This checks if the current codepoint is a multiple of 16 (remember that we added 1 to the codepoint in the top half of the loop). If it is a multiple of 16, then DP+ will turn the DP to face downwards. Otherwise, the DP stays facing left. This is based on the observation that all the characters before a newline all have codepoints that are 1 less than a multiple of 16.

For example, let's say that at the beginning of this portion, the stack is [48]. This means that the ascii 47, or /, just got printed, and then that value was incremented by 1 from the top half of the loop. After taking this value modulo 16 and applied ! on the value, the stack becomes [48,1]. The 1 on the stack is then multiplied by 3 which causes DP+ to turn the DP 3 times, resulting in the DP facing down.

Bottom branch of the loop:

     t
     ?
     kdvtrbbbbb

What it does:

Command                              Stack          DP,     CC
1 5 dup +                            [33,1,10]      left    right
out (char) pop                       [33]           left    right

Basically just prints out a newline.

Top branch of the loop:

     t    rr
     ?????rr?

Nothing actually happens here. The code just passes through and enters back into the loop. One thing to note is that the reason why we did CC+ earlier was to get through the r block. Before the CC+, the CC was facing right, but that poses a problem when we go through the r block, because the code will continue execution through the top left of the block, which will break the entire code. Instead, we toggle the CC once to switch it to the left, which causes the code to exit out of the block from the bottom left.

Once the top half is entered again, dup out (char) is ran, printing out another character, and the loop continues.

Loop exit:

                          s_
                           _

Once the loop reaches 127, the loop is exited. s is essentially a noop that does nothing, and the program terminates in the _ block.

Answer 6

R, 35 bytes

write(intToUtf8(32:126,T),1,16,,"")

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Answer 7

Rust, 58 bytes

||for i in' '..''{print!("{:
<1$}",i,i as usize%16/15+1)}

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Explanation

• The answer is a closure expression implementing the Fn() trait.

• print! can pad outputs to a given length: print!("{:x<5}", 'a') → axxxx

• The length can come from an argument: print!("{:x<1$}", 'a', 5) → axxxx

• The padding char can be a raw newline in place of the x in this example.

• In my program, the length is either 1 (just the char) or 2 (char + newline).

Answer 8

MS-DOS (.COM format), 33 30 26 bytes

• -3 thanks to Deadcode
• -4 by using INT 29H to output

Iterates through 20H and 7EH, printing each character and a line break between the 16th character of each line.

0000 B0 20 CD 29 40 3C 7F 74 10 A8 0F 75 F5 50 B0 0A
0010 CD 29 B0 0D CD 29 58 EB E9 C3

Assembly version (TASM):

IDEAL
P8086

MODEL   TINY
CODESEG
ORG 100H

MAIN:   
    MOV  AL,20H ; Start at ' '
PRINT:
    INT  29H    ; Print character
    INC  AX     ; Increment character
    CMP  AL,7FH ; Done?
    JZ   DONE   ; Yes, exit
    TEST AL,0FH ; End of line?
    JNZ  PRINT  ; No, next character
    PUSH AX     ; Save current character
    MOV  AL,0AH ; Print CR+LF
    INT  29H
    MOV  AL,0DH
    INT  29H
    POP  AX     ; Restore current character
    JMP  PRINT  ; Next character
DONE:
    RET

END MAIN
ENDS

Answer 9

Python 3, 71 bytes

-2 bytes thanks to @pxeger

for i in range(6):print(''.join(chr(j+i*16+32)for j in range(16-i//5)))

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Answer 10

Python 2, 66 bytes

for i in range(6):print bytearray(j+i*16+32for j in range(16-i/5))

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Based on @mousetail's solution; submitted as my own answer by their suggestion.

Answer 11

Rust, ^{74 72 70} 64 bytes

(' '..'').any(|i|print!("{i}")>if i as u8%16>14{print!("
")});

Text contains non-printable characters so might not render correctly. @Antiip saved me some bytes.

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Answer 12

Whitespace, 110 100 bytes

[S S S T    T   T   T   T   N
_Push_31][N
S S N
_Create_Label_LOOP][S S S T N
_Push_1][T  S S S _Add][S N
S _Duplicate][S S S T   T   T   T   T   T   T   N
_Push_127][T    S S T   _Subtract][N
T   S T N
_If_0_Jump_to_undefined_Label][S N
S _Duplicate][T N
S S _Print_as_character][S N
S _Duplicate][S S S T   S S S S N
_Push_16][T S T T   _Modulo][S S S T    T   T   T   N
_Push_15][T S S T   _Subtract][N
T   T   N
_If_neg_Jump_to_Label_LOOP][S S S T S T S N
_Push_10_\n][T  N
S S _Print_as_character][N
S N
N
_Jump_to_Label_LOOP]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Explanation in pseudo-code:

Integer n = 31
Start LOOP:
  n = n + 1
  If (n==127):
    Stop program with error by jumping to undefined Label
  Print n as character to STDOUT
  If (n modulo 16 < 15):
    Go to next iteration of LOOP
  Print '\n'
  Go to next iteration of LOOP

Answer 13

brainfuck, 64 86 bytes

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

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No longer prints 0x7f as the last character.

With many thanks to @Jiří for the fix.

Answer 14

x86-32 machine code (Linux executable) 27 bytes

cut-down NASM listing: Address | machine code | source

                   global _start
                   _start:            ; Linux processes (and thus static executables) start with registers zeroed, except ESP
                   asciitable:
 00 B020            mov al, ' '
 02 B20A            mov dl, 0xa
 04 89E7            mov edi, esp      ; space above ESP on the stack is argc, argv[] and env[] array elements, and the env strings.  We overwrite that.
                   .loop:
 06 AA              stosb
 07 40              inc  eax            ; there is no branch condition for AF, the half-carry flag :/
 08 A88F            test al, 0x8f       ; detect mod16 and when we've gone past printable
 0A 7503            jnz .nonewline
 0C 92              xchg  eax, edx
 0D AA              stosb
 0E 92              xchg  eax, edx
                   .nonewline:
                    ; FLAGS still set from earlier TEST
 0F 79F5            jns  .loop               ; stop at 0x80, beyond what the syscall uses
                     ; mov [edi-1], 0xa      ; no trailing newline
                   
 11 B004            mov  al, 4   ; __NR_write
                    ; ebx=0 from process startup; writing to stdin happens to work on a terminal
 13 89E1            mov  ecx, esp
 15 8D5060          lea  edx, [eax-4 + 0x7f-0x20 + 5]  ; exclusive range, 5 newlines; final line does not end with newline as whitespace isn't required
 18 CD80            int  0x80    ; write(0, ecx, 100)
                   
                   ; mov  eax,1      ; exit(ebx)
                   ; int  0x80
 1A CC              int3  ; abort program

Fill bytes into a buffer (overwriting stack memory above ESP, starting with argc, argv[], and into env[]). We go a bit beyond 0x7e because that makes the stop condition cheaper, but those bytes aren't printed: we have to generate an explicit length in EDX anyway, so we just omit it.

We make one write(0, buf, 100) system call. 0 is STDIN_FD, but on a normal xterm/konsole/gnome_terminal, FD 0,1, and 2 are all read+write duplicates of the same file description. So we save 1 byte for inc ebx. If you want to pipe the output into something to hexdump for example, ./asciitable 0>&1 | hexdump -C

Instead of existing cleanly, we use int3 to raise a debug exception, resulting in the OS delivering SIGTRAP, killing the process. A typical shell will then print Trace/breakpoint trap (core dumped), but that's the shell, not this program. This saves 3 bytes vs. mov al, 1 / int 0x80 (the upper bytes of EAX are zeroed, unless write returned an error).

I considered letting execution just fall off the end (to probably 00 00 padding added by the linker, which decodes as add [eax], al and will segfault), but we're already stretching things a bit by only counting the .text section of the executable, not the whole file size. (Unlike a DOS .com, there is metadata)

Our output doesn't end with a newline; as the question says, no trailing whitespace is required after the 0x7e ~ character.

The mod16 detection via checking the low bits 4 of AL is fairly straightforward, but was borrowed from ErikF's the x86-16 MS-DOS answer which I read first before thinking about how I'd do it.

But to enable a jns as the loop branch without a separate cmp, we also set the MSB in our mask for test. That's not a spot we wanted a newline anyway. I'd hoped to be able to branch on FLAGS from inc al (2 bytes), but since we'd also need something for newlines, this is even better: we can now use inc eax (1 byte) and still branch on the MSB of AL. Using mov dl, 0xa (2B) / xchg/stosb/xchg (1B each) instead of mov byte [edi], 0xa (3 bytes) / inc edi (1 byte but clobbers FLAGS) was the key to that, saving another test or cmp for a net saving of 1 byte.

Demo:

$ nasm -felf32 asciitable.asm
$ ld -melf_i386 -o asciitable asciitable.o
$ ./asciitable
 !"#$%&'()*+,-./
0123456789:;<=>?
@ABCDEFGHIJKLMNO
PQRSTUVWXYZ[\]^_
`abcdefghijklmno
pqrstuvwxyz{|}~Trace/breakpoint trap (core dumped)

$ ./asciitable 0>&1 | cat  # suppresses signal diagnostic from shell
 !"#$%&'()*+,-./
0123456789:;<=>?
@ABCDEFGHIJKLMNO
PQRSTUVWXYZ[\]^_
`abcdefghijklmno
pqrstuvwxyz{|}~peter@volta:/tmp$

$ ./asciitable 0>&1 | hexdump -C
00000000  20 21 22 23 24 25 26 27  28 29 2a 2b 2c 2d 2e 2f  | !"#$%&'()*+,-./|
00000010  0a 30 31 32 33 34 35 36  37 38 39 3a 3b 3c 3d 3e  |.0123456789:;<=>|
00000020  3f 0a 40 41 42 43 44 45  46 47 48 49 4a 4b 4c 4d  |?.@ABCDEFGHIJKLM|
00000030  4e 4f 0a 50 51 52 53 54  55 56 57 58 59 5a 5b 5c  |NO.PQRSTUVWXYZ[\|
00000040  5d 5e 5f 0a 60 61 62 63  64 65 66 67 68 69 6a 6b  |]^_.`abcdefghijk|
00000050  6c 6d 6e 6f 0a 70 71 72  73 74 75 76 77 78 79 7a  |lmno.pqrstuvwxyz|
00000060  7b 7c 7d 7e                                       |{|}~|
00000064

$ strace ./asciitable 0>/dev/null
execve("./asciitable", ["./asciitable"], 0x7ffc63d167c0 /* 55 vars */) = 0
[ Process PID=2670503 runs in 32 bit mode. ]
strace: WARNING: Proper structure decoding for this personality is not supported, please consider building strace with mpers support enabled.
write(0, " !\"#$%&'()*+,-./\n0123456789:;<=>"..., 100) = 100
--- SIGTRAP {si_signo=SIGTRAP, si_code=SI_KERNEL} ---
+++ killed by SIGTRAP (core dumped) +++
Trace/breakpoint trap (core dumped)

Try it online! with an inc ebx added so it writes to stdout and thus shows up in the tio.run output pane. IDK why it didn't show in the debug pane. (32-bit code supported by having FASM make an executable directly, instead of NASM.)

TODO: a function instead of program, filling a buffer

We'd obviously save the syscall code, but could no longer write past the end of the actual stop point. So would need maybe 1 more byte in the loop to go back to cmp al, 0x7e / jne as the loop condition.

Answer 15

x86-16 machine code, PC DOS, 24 19 bytes

00000000: b020 bb0a 0daa 40a8 0f75 0393 ab93 3c7e  . [email protected]....<~
00000010: 7ef3 c3                                  ~..

Listing:

B0 20           MOV   AL, 20H       ; starting codepoint
BB 0D0A         MOV   BX, 0D0AH     ; CR/LF chars
            CP_LOOP: 
AA              STOSB               ; write AL to buffer
40              INC   AX            ; increment to next char
A8 0F           TEST  AL, 0FH       ; is end of line?
75 03           JNZ   CP_NEXT       ; jump if so
93              XCHG  AX, BX        ; save current codepoint
AB              STOSW               ; write CR/LF to buffer
93              XCHG  AX, BX        ; restore codepoint
            CP_NEXT: 
3C 7E           CMP   AL, 07EH      ; <= 7EH? 
7E F3           JLE   CP_LOOP       ; if so, keep looping 
C3              RET                 ; return to caller

Using @PeterCordes's suggestion to write to an output buffer instead of direct to console.

Callable function, output to string buffer ES:[DI].

Answer 16

K (ngn/k), 21 17 bytes

-4bytes thanks to Traws!

`0:`c$6 16#32+!95

It's my first answer in K! So there might be some optimizations ^^

`0:`c$6 16#32+!95
           32+!95 / Creates an array from 32 to 126
      6 16#       / Reshapes it to 16 columns and 6 lines (16 can me replaced by 0N)
   `c$            / For each elements of this array/matrix, convert the int to a char
`0:               / Print the result whitout ( and "

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Answer 17

05AB1E, 10 bytes

Without builtin:

₃L31+çJ6ä»

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With builtin (5 bytes):

žQ6ä»

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Explanation:

₃L          # Push a list in the range [1,95]
  31+       # Add 31 to each
     ç      # Convert the codepoint-integers to ASCII characters
      J     # Join the list together to a string
       6ä   # Split it into 6 parts
         »  # Join by newlines
            # (after which the result is output implicitly)

žQ          # Push a string of all ASCII characters
  6ä»       # Same as above

Answer 18

Burlesque, 18 bytes

@!~r@' +]16co)++un

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@!~  # Push ! and ~ to stack
r@   # Range between them
' +] # Prepend " "
16co # Chunks of 16
)++  # Concat
un   # Intercalate newlines between blocks and join

Answer 19

PowerShell Core, 35 bytes

-join(' '..'~')-split'(.{16})'-ne''

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     (' '..'~')                     # create a list of characters from <space> to ~
-join                               # join the list to a single string
               -split'(.{16})'      # split the string every 16 chars ...
                              -ne'' # ... and let only non-empty strings pass.

Answer 20

BQN, 12 bytes

6‿16⥊' '+↕95

Try it at BQN REPL

Output has a trailing space character.

BQN's affine character space is quite neat here: adding a number to a character gives another character. So: create a range from zero to 95 (↕95), add it to the space character (' '+), and reshape to 6-row, 16-column format (6‿16⥊).

---

BQN, 26 bytes

@+100⥊⍉(32+⍉6‿16⥊↕95)∾6⥊10

Try it at BQN REPL

Longer version with output string that is truly "exactly like shown above", written before the edit allowing trailing whitespace.
Creates a transposed array of character values (32+⍉6‿16⥊↕95), appends a newline value 10 to each column (∾6⥊10), re-transposes (⍉) & extracts the first 100 values (100⥊), and finally converts to characters (@+).

Answer 21

V (vim), 10 bytes

¬ ~5ñ16|á

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Explanation:

¬               # Insert characters between the code points...
  ~             #   "space" and "~"
   5ñ           # 5 Times...
     16|        #   Go to the 16th character on this line
        á<cr>   #   Append a newline

Hexdump:

00000000: ac20 7e35 f131 367c e10d                 . ~5.16|..

Answer 22

Perl 5, 30 bytes

say map$/x!($_%16).chr,32..126

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Answer 23

sed -r, 94 bytes

s/.*/ABCDEFGHIJKLMNO\nPQRSTUVWXYZ/
saa !"#$%\&'()*+,-./\n0123456789:;<=>?\n@&[\\]^_\n`\L&{|}~a

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This is probably the best that can be done in sed, which has no functionality for ranges of characters. The main byte saving is in the fact we can generate ABCDEFGHIJKLMNOPQRSTUVWXYZ and then use \L to get the lowercase version.

Answer 24

PHP, 43 40 bytes

<?=chunk_split(join(range(' ','~')),16);

Simply outputs the expected text, without any trailing newlines.

Just run php file.php and it should produce the output.

Try it on: https://onlinephp.io/c/013ce

---

Thanks to Steffan for -3 bytes.

Answer 25

Vyxal 2.14.1, 9 bytes

kPð+s16ẇ⁋

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So turns out kP isn't actually all of printable ascii - it's missing the space character, and it isn't in the right order either, meaning that for the purposes of this challenge, it shouldn't be banned, as it doesn't print "the ascii table" and it isn't exactly "trivial" either.

(And if it isn't allowed, then that's grounds enough for VTCing this challenge as needs details or clarity, because "trivialising the challenge" isn't objectively defined - there's an argument to be made that kPð+s isn't a trivial built-in).

Explained

kPð+s16ẇ⁋
kPð+       # a string of 0-9a-Z + python's string.punctuation + space
    s      # sorted to be in printable ascii order
     16ẇ   # split into parts of length 16
        ⁋  # joined on newlines

Vyxal, 5 bytes

kP6/⁋

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Bug fixes make this 5 bytes in reality.

Answer 26

R, 38 bytes

intToUtf8(rbind(matrix(32:126,16),10))

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Answer 27

APL (Dyalog Extended), 12 bytes

6 16⍴' '…'~'

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' '…'~' inclusive range from space to tilde

6 16⍴ cyclically reshape into the required number of rows and columns

Answer 28

MathGolf, 10 bytes

♣⌡╒T≥$y☻/n

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Explanation:

♣           # Push 128
 ⌡          # Decrement it by 2 to 126
  ╒         # Pop and push a list in the range [1,126]
   T        # Push 31
    ≥       # Pop and remove the first 31 items from the list
     $      # Convert each codepoint-integer to a character
      y     # Join the list to a string
       ☻    # Push 16
        /   # Pop and split the string into parts of size 16
         n  # Join the list by newlines
            # (after which the entire stack is output implicitly as result)

Answer 29

jq -nr, 39 bytes

[range(32;127)]|implode|scan(".{1,16}")

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The best I can do without regex is 40 bytes:

16*range(2;8)|[.+range(16-./99)]|implode

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Answer 30

Charcoal, 12 bytes

ＵＴ⪪⪫…· ¦~ω¹⁶

Try it online! Link is to verbose version of code. Explanation:

ＵＴ

Disable automatic padding (would normally output a trailing space on the last line).

⪪⪫…· ¦~ω¹⁶

List the characters from to ~ inclusive, join them together, split into groups of 16, and implicitly print each group on its own line.

Save 6 bytes by using the builtin γ in place of ⪫…· ¦~ω. Try it online! Link is to verbose version of code.