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

C (clang), ^{51 \$\cdots\$ 49} 48 bytes

f(c){for(c=32;putchar(c)<126;++c%16||puts(""));}

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Saved a 2 bytes thanks to tsh!!!
Saved a bytes thanks to ovs!!!

Answer 2

Factor, 36 bytes

32 126 [a,b] 16 group "\n"join write

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Answer 3

Pyth, 15 9 bytes

jsMc6rd\

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There's a literal DEL character after the \.

jsMc6rd\
     rd\  Half-inclusive string range from ' ' (space) to DEL
   c6     Chop into 6 equal-sized lists (last chunk is smaller)
 sM       Join each list on '' (empty)
j         Join on newlines, implicitly output

Answer 4

Raku, 33 bytes

.say for (' '..'~').join.comb(16)

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Answer 5

GeoGebra, 68 67 bytes

UnicodeToText(Flatten(Zip(If(Mod(a,16)>14,{a,10},{a}),a,32...126)))

Text is shown on the graph.

Might be golfable with some different strategy, but I'm not too experienced with GeoGebra.

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Answer 6

Ruby, 53 44 bytes

Thanks user for telling me about each_slice, saving 9 bytes!

(' '..'~').each_slice(16).each{|x|puts x*""}

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My first time golfing in Ruby lol, just wanted to try a new language :D . Tell me if there are any golfs, I'm anticipating a lot!

Answer 7

stacked, 23 bytes

95:>32+16 chunk chr out

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Slightly shorter than the more obvious 24 bytes:

32 126|>16 chunk chr out

Explanation

95:>32+16 chunk chr out
95:>                       the range of numbers from 0 to 94 inclusive
    32+                    add 32 to each
       16 chunk            split into chunks of 16
                chr        convert each number to a character
                    out    output it

stacked's default output behavior with out for 2D character arrays is exactly joining them, and produces the correct output.

Answer 8

Pip, 15 bytes

P*C:32,127<>16u

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Explanation

P*C:32,127<>16u
    32,127       Range(32, 127)
          <>16   Group into chunks of size 16 (with the last one being shorter)
  C:             Convert to characters
P*               Print each sublist (concatenated together, by default)
              u  Suppress autoprinting of the last expression

Or, with the -l flag to handle the output formatting, it would be 12 bytes:

C:32,127<>16

Using built-in

PA is a string containing all printable ASCII characters, so:

PA<>16Jn

Or, with the -l flag:

PA<>16

Answer 9

Regenerate -a -s=, 45 bytes

[ -/]|
|[0-?]|
|[@-O]|
|[P-_]|
|[`-o]|
|[p-~]

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Explanation

The -a flag requests all possible matches; the -s= flag sets the separator between matches to empty string. Then the regex is just a big alternation: "match any character from to /, or a newline, or any character from 0 to ?, or a newline, or..." The regex always matches one character, and the -a flag outputs all of them.

Answer 10

Brainfuck + head + sh, 65 + 10 bytes

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

Execute <brainfuck-program>|head -c-2 (see meta).

head removes a trailing \del char of the brainfuck program output (as noted by @mousetail).

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Answer 11

Kotlin, 80 72 56 bytes

56 bytes versuin

(' '..'~').chunked(16).flatMap{it+"\n"}.joinToString("")

Directly using a char range instead of an integer range.

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72 bytes version

(32..126).chunked(16).flatMap{it.map{it.toChar()}+"\n"}.joinToString("")

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80 bytes version

(32..126).chunked(16){it.joinToString(""){""+it.toChar()}+"\n"}.joinToString("")

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Answer 12

Nibbles, 7.5 bytes (15 nibbles)

`/16+' '`,95

`/16+' '`,95
        `,95    # range from 0 to 95
    +' '        # add each to a space character
`/16            # and split into chunks of 16

---

Nibbles, 5 bytes (10 nibbles)

`/16>>`<@

Uses the special value assigned to @ when input is empty, corresponding to a list of ascii characters including newline, in a non-standard order.

`/16>>`<@
        @       # " abcdefghijklmnopqrstuvwxyz.,!?_"
                # + newline +
                # "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789-+:;\"'~`@#$%^&*()[]{}<>\\/=|"
      `<        # sort 
    >>          # remove first item (newline)
`/16            # and split into chunks of 16
```

Answer 13

Piet + ascii-piet, 59 bytes (4×26=104 codels)

tliusqrfvrtqqqqijsmva???Bt  t?sd???vmjftrqaaaaqrsb_t eTt ee

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How it works

4 dup * dup +         Initialize n to 32
Loop:
    dup outC          [n] Print as char
    1 + dup dup       [n n n] Add 1 and make two more copies
    4 dup * % !       [n n n%16==0] Is this number divisible by 16?
    CC+               [n n] If so,
        5 dup + outC        Print newline
    5 dup dup * * >   [n n>125] Is this number greater than 125?
    ! DP+             [n] If not, go back to start of Loop
outC                  Otherwise, print n (126) and exit

Answer 14

Clojure, 72 63 bytes

(doseq[x(range 32 127)](print(str(char x)({15 \
}(rem x 16)))))

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Answer 15

Retina, 19 bytes

6*$(16*1¶
Y`1` -~_

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6*$(16*1¶

Insert 6 rows of 16 1s.

Y`1` -~_

Cyclically transliterate the first 95 1s to the range from to ~ and delete the 96th.

Answer 16

Husk, 11 bytes

↓2C16mcŀ127

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       ŀ127 # range from zero to 126
     mc     # map across this & get characters
  C16       # cut into sublists with length 16
↓2          # drop the first 2 of them

Answer 17

Python 2, 68 bytes

l=''.join(chr(x+32)for x in range(95))
while l:print l[:16];l=l[16:]

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Answer 18

MATL, 8 bytes

'~':16e!

Try it online! Outputs a trailing whitespace (to make the last row have the same length as the others).

How it works

'~'   % Push this character
:     % Range from space to that character, including both. Gives a character vector
16e   % Format as a 16-row character matrix, in major-column order. Pads with
      % character 0 at the end
!     % Transpose
      % Implicit display. Character 0 is displayed as space

Answer 19

Jelly, 10 bytes

32r126Ọs⁴Y

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5 if we can just use our langauage's features:

ØṖs⁴Y

How?

32r126Ọs⁴Y - Link: no arguments
32         - 32                }
   126     - 126               }
  r        - inclusive range   }
      Ọ    - cast to ordinals  } - together this is the same as ØṖ yields
        ⁴  - 16
       s   - split into chunks of length (up to 16)
         Y - join with newline characters

Answer 20

Bubblegum, 102 bytes

00000000: 5350 5452 5651 5553 d7d0 d4d2 d6d1 d5d3  SPTRVQUS........
00000010: e732 3034 3236 3135 33b7 b0b4 b2b6 b1b5  .20426153.......
00000020: b3e7 7270 7472 7671 7573 f7f0 f4f2 f6f1  ..rptrvqus......
00000030: f5f3 e70a 080c 0a0e 090d 0b8f 888c 8a8e  ................
00000040: 898d 8be7 4a48 4c4a 4e49 4d4b cfc8 ccca  ....JHLJNIMK....
00000050: cec9 cdcb e72a 282c 2a2e 292d 2baf a8ac  .....*(,*.)-+...
00000060: aaae a9ad 0300                           ......

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For some reason, compressing it made it actually a bit larger than the original string.

Answer 21

JavaScript (Node.js), 63 bytes

for(i=33,a=' ';i<129;a+=Buffer([i++]))i%16||console.log(a,a='')

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Answer 22

Haskell, 65 bytes

_#[]=[]
n#x=take n x:n#drop n x
main=putStr$unlines$16#[' '..'~']

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Answer 23

Forth (gforth), 52 bytes

: f 127 32 do i emit i 16 mod 15 = if cr then loop ;

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Explanation

Loops from 32 to 127 (end is non-inclusive). Outputs the ascii char for each number, if index % 16 == 15, outputs a newline.

Code Explanation

: f            \ start a new word definition
  127 32       \ set up loop bounds
  do           \ start a new counted loop
    i emit     \ output the ascii character for the current loop index
    i 16 mod   \ get index modulo 16
    15 =       \ check if result is equal to 15
    if cr then \ if it is, output a newline
  loop         \ end the counted loop
 ;             \ end the word definition

Answer 24

Acc!!, 67 bytes

Count i while i-100 {
1/((i+1)%17+1)
Write 10*_+(i+32-i/17)*(1-_)
}

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Explanation

We run a single loop that outputs one character each time, using math to switch between ASCII characters and newlines.

The loop index i runs from 0 through 99. We want to print newlines at indices 16, 33, 50, 67, and 84--that is, where (i+1)%17 is zero. With no comparison operators in Acc!!, we use integer division instead: when (i+1)%17 is zero, 1/((i+1)%17+1) is one, and it is zero otherwise. We store this quantity in the accumulator.

Now if the accumulator value is one, we want to write a newline, and if it is zero, we want to write something else. This is easily accomplished by multiplying the two quantities by _ and (1-_) respectively, and adding the results: Write 10*_+(...)*(1-_). It only remains to calculate the ASCII character's codepoint. At first, it is just i+32, but we need to adjust for the newlines. The number of newlines printed so far is i/17, so subtracting this from the index gives us the correct codepoint formula: i+32-i/17.

---

A more straightforward solution with two loops is 73 bytes:

Count i while i-6 {
Count j while j-16+i/5 {
Write i*16+32+j
}
Write 10
}

Answer 25

*><>, 53 bytes

e9*&1f+2*   v
v;?=&:&:+1o:<
\:1f+%?!`   /
        `ao

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Unfortunately, I feel like a lot of these bytes are spent setting, storing, and checking if we're at 126 yet, but I'm unsure of a better way of accomplishing this.

e9*&            push 126 to the stack, and pop it into a register
    1f+2*       push 32 to the stack
            v   Change IP direction downwards

            <   Change IP direction to the left
          o:    Duplicate the top of the stack, then pop it to STDOUT as a char
       :+1      Add one to the value on the top of the stack, and duplicate it
    &:&         Pop the register back onto the stack, duplicate it, then pop that into the register
 ;?=            If our two duplicated values match, end execution
v               Change IP direction downwards

\               Mirror IP to the right
 :              Duplicate the top of the stack
  1f+           Put 16 on the stack
     %          Get the modulus of our counter and 16
      ?!`       If the result is 0, drop down a line, and 
         ao         Print a newline
        `           And raise back up a line
            /   Else, mirror the IP up

Answer 26

PARI/GP, 43 bytes

for(i=1,95,printf("%c",i+31);i%16||print())

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---

PARI/GP, 44 bytes

for(i=2,7,print(Strchr([s=16*i..s+15-i\7])))

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Answer 27

Fig, \$17\log_{256}(96)\approx\$ 13.993 bytes

Without printable ascii builtin

nOC+32r95 16'+xcn

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With printable ascii builtin \$10\log_{256}(96)\approx\$ 8.231 bytes

ncp16'+xcn

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nOC+32r95 16'+xcn
      r95          # range [0,95)
   +32             # add 32 to each, range [32,126)
  C                # convert each element to str
 O                 # join by nothing
n         16       # every 16th element, apply the function
            '+     # add
               cn  # newline
              x    # to the element

ncp16'+xcn
 cp                # printable ascii
n  16'+xcn         # same as above

Answer 28

Julia 1.0, 53 52 bytes

0:16:80 .|>n->(prod(' ':'~')*" ")[n+1:n+16]|>println

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Answer 29

Julia 1.0, 69 63 bytes

map(i->println(String(reshape([' ':'~';' '],(16,6))[:,i])),1:6)

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-6 bytes thanks to @amelies: ' ':'~' is a valid range in Julia; no need to convert from integers

Answer 30

><>, 30 29 bytes

84*v
%?!\:e9*)?;:o1+:82*
ao \

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