100
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In this challenge, you should write a program or function which takes no input and prints or returns a string with the same number of bytes as the program itself. There are a few rules:

  • You may only output bytes in the printable ASCII range (0x20 to 0x7E, inclusive), or newlines (0x0A or 0x0D).
  • Your code must not be a quine, so the code and the output must differ in at least one byte.
  • Your code must be at least one byte long.
  • If your output contains trailing newlines, those are part of the byte count.
  • If your code requires non-standard command-line flags, count them as usual (i.e. by adding the difference to a standard invocation of your language's implementation to the byte count), and the output's length must match your solution's score. E.g. if your program is ab and requires the non-standard flag -n (we'll assume it can't be combined with standard flags, so it's 3 bytes), you should output 5 bytes in total.
  • The output doesn't always have to be the same, as long as you can show that every possible output satisfies the above requirements.
  • Usual quine rules don't apply. You may read the source code or its size, but I doubt this will be shorter than hardcoding it in most languages.

You may write a program or a function and use any of the standard methods of providing output. Note that if you print the result, you may choose to print it either to the standard output or the standard error stream, but only one of them counts.

You may use any programming language, but note that these loopholes are forbidden by default.

This is , so the shortest valid answer – measured in bytes – wins.

Leaderboard

var QUESTION_ID=121056,OVERRIDE_USER=8478;function answersUrl(e){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}function commentUrl(e,s){return"https://api.stackexchange.com/2.2/answers/"+s.join(";")+"/comments?page="+e+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+COMMENT_FILTER}function getAnswers(){jQuery.ajax({url:answersUrl(answer_page++),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){answers.push.apply(answers,e.items),answers_hash=[],answer_ids=[],e.items.forEach(function(e){e.comments=[];var s=+e.share_link.match(/\d+/);answer_ids.push(s),answers_hash[s]=e}),e.has_more||(more_answers=!1),comment_page=1,getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:!0,success:function(e){e.items.forEach(function(e){e.owner.user_id===OVERRIDE_USER&&answers_hash[e.post_id].comments.push(e)}),e.has_more?getComments():more_answers?getAnswers():process()}})}function getAuthorName(e){return e.owner.display_name}function process(){var e=[];answers.forEach(function(s){var r=s.body;s.comments.forEach(function(e){OVERRIDE_REG.test(e.body)&&(r="<h1>"+e.body.replace(OVERRIDE_REG,"")+"</h1>")});var a=r.match(SCORE_REG);a&&e.push({user:getAuthorName(s),size:+a[2],language:a[1],link:s.share_link})}),e.sort(function(e,s){var r=e.size,a=s.size;return r-a});var s={},r=1,a=null,n=1;e.forEach(function(e){e.size!=a&&(n=r),a=e.size,++r;var t=jQuery("#answer-template").html();t=t.replace("{{PLACE}}",n+".").replace("{{NAME}}",e.user).replace("{{LANGUAGE}}",e.language).replace("{{SIZE}}",e.size).replace("{{LINK}}",e.link),t=jQuery(t),jQuery("#answers").append(t);var o=e.language;/<a/.test(o)&&(o=jQuery(o).text()),s[o]=s[o]||{lang:e.language,user:e.user,size:e.size,link:e.link}});var t=[];for(var o in s)s.hasOwnProperty(o)&&t.push(s[o]);t.sort(function(e,s){var F=function(a){return a.lang.replace(/<\/?a.*?>/g,"").toLowerCase()},el=F(e),sl=F(s);return el>sl?1:el<sl?-1:0});for(var c=0;c<t.length;++c){var i=jQuery("#language-template").html(),o=t[c];i=i.replace("{{LANGUAGE}}",o.lang).replace("{{NAME}}",o.user).replace("{{SIZE}}",o.size).replace("{{LINK}}",o.link),i=jQuery(i),jQuery("#languages").append(i)}}var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe",COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk",answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;getAnswers();var SCORE_REG=/<h\d>\s*([^\n,]*[^\s,]),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/,OVERRIDE_REG=/^Override\s*header:\s*/i;
body{text-align:left!important}#answer-list,#language-list{padding:10px;width:290px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr></thead> <tbody id="answers"> </tbody> </table> </div><div id="language-list"> <h2>Winners by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr></thead> <tbody id="languages"> </tbody> </table> </div><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr></tbody> </table>

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  • 2
    \$\begingroup\$ Related. Related. \$\endgroup\$ – Martin Ender May 17 '17 at 11:19
  • 20
    \$\begingroup\$ "Your code must not be a quine" but... but... it's tagged quine \$\endgroup\$ – Okx May 17 '17 at 11:21
  • 4
    \$\begingroup\$ @Okx Because it's a generalised quine, i.e. the required output depends on the source code. \$\endgroup\$ – Martin Ender May 17 '17 at 11:22
  • 4
    \$\begingroup\$ @MartinEnder You should probably disallow output by exit code, which is a default. If you allow it nearly every one byte program in nearly every language is allowed. One user has already done this \$\endgroup\$ – Ad Hoc Garf Hunter May 17 '17 at 22:37
  • 2
    \$\begingroup\$ @WheatWizard output by exit code is not a string, so it doesn't apply here. \$\endgroup\$ – Martin Ender May 18 '17 at 4:29

240 Answers 240

1
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1
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Chip, 78+3 = 81 bytes

Flag: -w

Code (Try it online!):

g*
,xZ.
`@'|
,xZ<
`@'|
,xZ<
`@'|
,xZ<
`@'`.
,xZ~<
`@','
,xZ^.
`@'t{*
,xZ~'
`@'

Out: @@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@@

This uses a binary counter to halt at the correct time.


11+3 = 14 bytes (uses version string)

Flag: -V

Code (anything will do, just need to fill the length):

gibberishes

Out (current interpreter, has trailing newline):
chip.py 0.1.2


40+3 = 43 bytes (error message, uses stderr)

Flag: -w

Code (apparently some of this is filler?):

!*T :Hello you! That shirt looks great!;

Out: 1:1 WARN: '!' (33) is not a valid character

| improve this answer | |
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1
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JavaScript, 35 bytes

Source:

console.log(([]+[])["constructor"])

Output:

function String() { [native code] }
| improve this answer | |
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1
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Add++, 2 bytes

Noncompeting as language postdates challenge.

O

Try it online!

Outputs a 0 with a trailing newline.

| improve this answer | |
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1
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C#, 53 bytes

Not short, but good enough:

string a="1";for(int i=0;i<53;i++){Console.Write(a);}

Some Facts I can't resist to post:

  • Fifty-three is the 16th prime number. It is also an Eisenstein prime, and a Sophie Germain prime.

  • 53 cannot be expressed as the sum of any integer and its base-10 digits, making 53 a self number.

  • 53 is the smallest prime number that does not divide the order of any sporadic group.

| improve this answer | |
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1
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C#, 12 bytes

()=>1e11+"";

It returns 10^11, which is a 1 followed by 11 0-s, so 12 bytes long.

| improve this answer | |
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1
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cQuents, 5 bytes

#3::$

Outputs 1,2,3. This works because of the mode, ::. :: prints the sequence up to n, which in this case is hardcoded in as 3. $ prints the current index - it could be replaced with any single digit. So, the interpreter prints out the first three items in the $ sequence, separated by the default delimiter, ,.

| improve this answer | |
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1
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Common Lisp, 5 bytes

1e+01

in the Common Lisp REPL it produces 10.0 plus linefeed, for a total of 5 bytes.

Thanks to @MartinEnder for noting an error in the previous 4 bytes version!

| improve this answer | |
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1
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PowerShell, 5 bytes

$true

Output (including trailing linefeed):

True

Try it online!

| improve this answer | |
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1
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Triangular, 9 bytes

9.(,%<>d]

Try it online!

Formats into this triangle:

    9 
   . ( 
  , % < 
 > d ] ÿ

Triangular auto-inserts ÿ wherever there is no source code to fill the smallest triangle.

Without control flow, the program looks like 9(%d]. Explanation:

  • 9 - push 9 to the stack
  • ( - open loop
  • % - print top of stack as integer
  • d - decrement top of stack
  • ] - jump back to ( if top of stack is truthy
| improve this answer | |
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1
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Cubically, 2 bytes

%5

Try it online! Boring. Outputs 45. Also works with %4 (36), %3 (27), and %2 (18).

Cubically, 7 bytes

+5*66%6

Try it online! Outputs 4100625 by adding 45 to the notepad, then squaring it twice.

Cubically, 9 bytes

+5*5555%6

Try it online! Outputs 184528125 by adding 45 to the notepad, then multiplying it by 45 four times.

| improve this answer | |
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1
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Python 3, 10 bytes

print(1e6)

Try it online!

| improve this answer | |
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1
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Common Lisp REPL, 2 bytes

'a

returns a followed by a newline

| improve this answer | |
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1
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Cubically, 2 bytes

%5

Try it online!

Prints the sum of all values on face 5. Face 5 is initialized to

555
555
555

so this prints

45

(Other valid answers: %2,%3,%4)

| improve this answer | |
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  • \$\begingroup\$ Note \$\endgroup\$ – MD XF Sep 12 '17 at 3:51
  • 1
    \$\begingroup\$ @MDXF I really need to start reading existing answers more thoroughly instead of just assuming they haven't been solved in Cubically \$\endgroup\$ – Kamil Drakari Sep 12 '17 at 13:04
1
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Recursiva, 2 bytes

L(

Try it online!

It outputs 26.

L(
L   - Length of
 (  - upper-case alphabet yield
| improve this answer | |
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1
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Implicit, 1 byte

#

Pushes the length of the stack to the stack. Implicit output. Try it online!

1 byte alternatives

  • ß prints a space.
  • ±, $, +, -, *, /, _, and ^ all push 0 if the TIO input box is empty.
| improve this answer | |
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1
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SmileBASIC, 3 bytes

?#Y

Output:

128

#Y is a constant used for checking the (Y) button, and has a value of 128.

| improve this answer | |
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  • \$\begingroup\$ I wonder if ? is acceptable since it prints a blank line. (Of course it's impossible to actually verify if the console writes a linefeed.) \$\endgroup\$ – snail_ Apr 20 '18 at 3:47
1
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x86 .COM opcode, 9 Bytes

0100 B409          MOV     AH,09
0102 BAFF00        MOV     DX,00FF
0105 CD21          INT     21
0107 C3            RET
0108 24            DB      '$'
| improve this answer | |
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1
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Pascal (FPC), 22 bytes

begin write('':22)end.

Try it online!

:22 pads the write value to be at least 22 characters in length. Here, the write value is an empty string, so this prints 22 spaces.

| improve this answer | |
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1
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Flobnar, 9 8 bytes

*<*@
9*<

Try it online!

Outputs 43046721, which is 98. This uses the -d flag to output in decimal. Thanks to Esolanging Fruit for this solution

| improve this answer | |
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1
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Z80Golf, 5 3 bytes

00000000: 3e76 34                                  >v4

Try it online!

The output is vvv. Yay for ASCII-printable machine code!

Disassembly

start:
  ld a, $76 ; 3e 76
  inc (hl)  ; 34

Uses the same concept as the 5-byte one: edit the running code on the fly, and abuse stack underflow.

The program initially loads 'v' to register a, and increments the value at the memory address (hl), which is $0000 where the code $3e is located. Then the code becomes:

  ccf      ; 3f ; Complement carry flag, effective no-op in this program
  halt     ; 76
  inc (hl) ; 34

So the halt is uncovered right away. The stack underflow magic takes the next job; putchar's ret returns to $763f, $0034, and $0000 in the order, and three vs are printed in the process. Now pc is back at the start of the program. ccf is no-op, halt is executed, and the program terminates.


A variation, 4 bytes

00000000: 3e76 343b                                >v4;

Try it online!

The output is vvvv.

Disassembly

start:
  ld a, $76 ; 3e 76
  inc (hl)  ; 34
  dec sp    ; 3b

The added dec sp makes things slightly more convoluted; the stack is arranged so that putchar is run 4 times instead of 3. Without the instruction, the return addresses are $763f - $3b34 - $0000; with it, the addresses are $3e00 - $3476 - $003b - $0000.

Leaving this solution here, in case someone finds the "stack misalignment" technique useful.


Previous solution, 5 bytes

00000000: 2e0a 3e76 34                             ..>v4

Try it online!

Disassembly

start:
  ld l, $0a  ; 2e 0a
  ld a, $76  ; 3e 76
  inc (hl)   ; 34

The second instruction sets up the character to print, which is 'v'. The rest increases the value at the memory address $000a.

The output is vvvvv. Too bad it's not in uppercase (and it's not six v's), otherwise I'd reference the game VVVVVV.

How it works

For no-input challenges, underflowing the stack into the code is a standard technique in Z80Golf. It is done by letting the PC flow beyond the end of the code, so that putchar at address $8000 is reached, a char is printed, and ret is executed.

  • When the code section is run the first time, the instruction at memory $000a becomes ld bc, $0000 (opcode $01 $00 $00). Pretty much no-op, since all registers are zeroed when the program starts. putchar is reached and v is printed once. Then it returns to $0a2e.
  • putchar is run again, v is printed, and the next return address is $763e.
  • Same thing again, return to $0034.
  • Same thing again, finally return to $0000, the start of the code.
  • The code is run again; the instruction at memory $000a is now ld (bc), a (opcode $02). a is $76, so the value is written to memory $0000. After putchar is reached this time, return to $0000 again, and PC meets the new instruction $76 which is... halt! The program ends.
| improve this answer | |
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1
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MathGolf, 1 byte

!

Try it online!

For MathGolf, there is an implicit pop if nothing is on the stack and nothing is in the input. The operator will pop the default value for the type it's requesting, which is either 0, [] or "". That means that a lot of operators will pop implicit zeros from the empty stack, and transform them either into 0 or 1. There are 15 1-byte programs which satisfy the criteria of this challenge, most of them output 0. The rest output 1.

Output 0

*: 0
+: 0
,: 0
-: 0
.: 0
<: 0
>: 0
f: 0
i: 0
w: 0
x: 0

Output 1

!: 1
#: 1
): 1
=: 1
| improve this answer | |
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  • \$\begingroup\$ Doesn't ! output 1 as well by default? I remember using that in the How high can you count challenge. Also, your code contains *, but your TIO is . \$\endgroup\$ – Kevin Cruijssen Dec 6 '18 at 12:04
  • \$\begingroup\$ You're absolutely correct. I created a script to search for all possible correct answers, ! must have gotten lost somewhere. My first thought was , but that's not printable ascii. I'll update the answer once I've looked for more possible correct answers. \$\endgroup\$ – maxb Dec 6 '18 at 12:06
1
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PowerShell, 4 bytes

1..2

Try it online!

Output includes a trailing newline!

| improve this answer | |
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  • \$\begingroup\$ You could to try ,1 or +1. Output includes a trailing newline! \$\endgroup\$ – mazzy Dec 9 '18 at 7:04
  • \$\begingroup\$ Thanks @mazzy, however I think that this method has already been used in this answer. \$\endgroup\$ – Gabriel Mills Dec 9 '18 at 15:21
1
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Javascript (ES6), 5 bytes

!1+''

outputs the string "false".

| improve this answer | |
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1
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Perl6 12 7 bytes

say |^6

Prints the number from 0-5 and a newline for 7 bytes. Suggested by Jo King.

Original

.say for ^7
| improve this answer | |
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  • \$\begingroup\$ Nice. I'll have that :D (With attribution of course). \$\endgroup\$ – Scimon Proctor Jul 29 '19 at 13:34
1
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Alchemist, 9 bytes

_->9Out__

Try it online!

Uses up the initial _ atom to output the number of _ atoms remaining (0) nine times.

| improve this answer | |
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1
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brainfuck, 24 bytes

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

Try it online!

Prints 24 Us.

Explanation:

-[>+<---]    # Generate a U
++++++[      # Loop 6 time
->....<]     # Printing U 4 times each
| improve this answer | |
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1
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Keg, 2 bytes

Q.

Outputs 81, which is the ascii value of Q

Try it Online!

| improve this answer | |
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  • 2
    \$\begingroup\$ I thought ! would be shorter, but the implicit output has a trailing newline :( \$\endgroup\$ – Jo King Aug 10 '19 at 9:55
  • 1
    \$\begingroup\$ It also has 3 trailing spaces for some reason \$\endgroup\$ – EdgyNerd Aug 10 '19 at 10:01
1
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T-SQL, 21 15 bytes

print space(15)

Try it online

Output:

               
| improve this answer | |
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  • \$\begingroup\$ Looks like a space is a valid output, so PRINT SPACE(15) would save you some bytes. \$\endgroup\$ – BradC Aug 14 '19 at 15:12
  • \$\begingroup\$ Nice! Thanks... \$\endgroup\$ – mbomb007 Aug 14 '19 at 15:16
1
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Pyramid Scheme, 32 bytes

^ ^ ^
-^-^-
^-^-
-^-^
^-^-
- -


Try it online!

Outputs 16 zeroes and newlines. I can find a lot of alternative 32 byters with varying amounts of extra padding, but the closest I can come to something shorter is this 27 byte program that outputs 26 bytes

| improve this answer | |
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1
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Keg, 2 bytes

2-byte answer:


Note that there is a 0x01 here. Try It Online!

| improve this answer | |
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