Using your language of choice, golf a quine.

A quine is a non-empty computer program which takes no input and produces a copy of its own source code as its only output.

No cheating -- that means that you can't just read the source file and print it. Also, in many languages, an empty file is also a quine: that isn't considered a legit quine either.

No error quines -- there is already a separate challenge for error quines.

Points for:

  • Smallest code (in bytes)
  • Most obfuscated/obscure solution
  • Using esoteric/obscure languages
  • Successfully using languages that are difficult to golf in

The following Stack Snippet can be used to get a quick view of the current score in each language, and thus to know which languages have existing answers and what sort of target you have to beat:


var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk";var answers=[],answers_hash,answer_ids,answer_page=1,more_answers=!0,comment_page;function answersUrl(index){return"https://api.stackexchange.com/2.2/questions/"+QUESTION_ID+"/answers?page="+index+"&pagesize=100&order=desc&sort=creation&site=codegolf&filter="+ANSWER_FILTER}
function commentUrl(index,answers){return"https://api.stackexchange.com/2.2/answers/"+answers.join(';')+"/comments?page="+index+"&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(data){answers.push.apply(answers,data.items);answers_hash=[];answer_ids=[];data.items.forEach(function(a){a.comments=[];var id=+a.share_link.match(/\d+/);answer_ids.push(id);answers_hash[id]=a});if(!data.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(data){data.items.forEach(function(c){if(c.owner.user_id===OVERRIDE_USER)
answers_hash[c.post_id].comments.push(c)});if(data.has_more)getComments();else if(more_answers)getAnswers();else process()}})}
getAnswers();var SCORE_REG=(function(){var headerTag=String.raw `h\d`
var score=String.raw `\-?\d+\.?\d*`
var normalText=String.raw `[^\n<>]*`
var strikethrough=String.raw `<s>${normalText}</s>|<strike>${normalText}</strike>|<del>${normalText}</del>`
var noDigitText=String.raw `[^\n\d<>]*`
var htmlTag=String.raw `<[^\n<>]+>`
return new RegExp(String.raw `<${headerTag}>`+String.raw `\s*([^\n,]*[^\s,]),.*?`+String.raw `(${score})`+String.raw `(?=`+String.raw `${noDigitText}`+String.raw `(?:(?:${strikethrough}|${htmlTag})${noDigitText})*`+String.raw `</${headerTag}>`+String.raw `)`)})();var OVERRIDE_REG=/^Override\s*header:\s*/i;function getAuthorName(a){return a.owner.display_name}
function process(){var valid=[];answers.forEach(function(a){var body=a.body;a.comments.forEach(function(c){if(OVERRIDE_REG.test(c.body))
body='<h1>'+c.body.replace(OVERRIDE_REG,'')+'</h1>'});var match=body.match(SCORE_REG);if(match)
valid.push({user:getAuthorName(a),size:+match[2],language:match[1],link:a.share_link,})});valid.sort(function(a,b){var aB=a.size,bB=b.size;return aB-bB});var languages={};var place=1;var lastSize=null;var lastPlace=1;valid.forEach(function(a){if(a.size!=lastSize)
lastPlace=place;lastSize=a.size;++place;var answer=jQuery("#answer-template").html();answer=answer.replace("{{PLACE}}",lastPlace+".").replace("{{NAME}}",a.user).replace("{{LANGUAGE}}",a.language).replace("{{SIZE}}",a.size).replace("{{LINK}}",a.link);answer=jQuery(answer);jQuery("#answers").append(answer);var lang=a.language;lang=jQuery('<i>'+a.language+'</i>').text().toLowerCase();languages[lang]=languages[lang]||{lang:a.language,user:a.user,size:a.size,link:a.link,uniq:lang}});var langs=[];for(var lang in languages)
langs.push(languages[lang]);langs.sort(function(a,b){if(a.uniq>b.uniq)return 1;if(a.uniq<b.uniq)return-1;return 0});for(var i=0;i<langs.length;++i)
{var language=jQuery("#language-template").html();var lang=langs[i];language=language.replace("{{LANGUAGE}}",lang.lang).replace("{{NAME}}",lang.user).replace("{{SIZE}}",lang.size).replace("{{LINK}}",lang.link);language=jQuery(language);jQuery("#languages").append(language)}}
body{text-align:left!important}#answer-list{padding:10px;float:left}#language-list{padding:10px;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="https://cdn.sstatic.net/Sites/codegolf/primary.css?v=f52df912b654"> <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><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><table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td><a href="{{LINK}}">{{SIZE}}</a></td></tr></tbody> </table> 

  • 4
    \$\begingroup\$ Do you not mean, "Golf you a quine for greater good!"? \$\endgroup\$ May 3, 2011 at 2:49
  • 62
    \$\begingroup\$ @muntoo it's a play on "Learn you a Haskell for Great Good". \$\endgroup\$ May 3, 2011 at 2:52
  • 18
    \$\begingroup\$ Did anybody notice that this is question 69? \$\endgroup\$
    – aidan0626
    Oct 24, 2020 at 22:47

417 Answers 417

1 2
4 5

Bubblegum, 105 Bytes


00000000: 0000 00ff ff00 0000 ffff 0000 00ff ff00  ................
00000010: 1400 ebff 0000 00ff ff00 0000 ffff 0000  ................
00000020: 00ff ff00 1400 ebff 4288 21c4 0000 1400  ........B.!.....
00000030: ebff 4288 21c4 0000 1400 ebff 4288 21c4  ..B.!.......B.!.
00000040: 0000 1400 ebff 4288 21c4 0000 1400 ebff  ......B.!.......
00000050: 0000 00ff ff00 0000 ffff 0000 00ff ff03  ................
00000060: 1300 0000 0313 0000 00                   .........

Try it online!

(You might want to verify it offline - since the input is hexdump and the output is raw.)

This relies on the fact that Bubbleugum tries to DEFLATE decode its input first:

o = zlib.decompress(code, -zlib.MAX_WBITS)

So if we can find a fixpoint in DEFLATE compression, such that x = zlib.decompress(x, -zlib.MAX_WBITS), we are done. But how to do this?

Part I: Generic Compression Quine

Say we have a compression programming 'language' that has two operations:

  • Pn: Print the following n tokens as literals, and skip interpreting them
  • Rn: Print the last n tokens printed

Let's write some simple programs in this to understand how it works.

Input   | Output
P1 P0   | P0
Input   | Output
P1 P0   | P0
P1 P1   | P1
Input   | Output
P1 P0   | P0
R1      | P0
Input          | Output
P4 P0 P0 P0 P0 | P0 P0 P0 P0
R4             | P0 P0 P0 P0

Now the question is: Just with these two instructions, can we create a quine? The answer is yes, thanks to Russ Cox:

Input          | Output
P0             | 
P0             |
P0             |
P4 P0 P0 P0 P4 | P0 P0 P0 P4
R4             | P0 P0 P0 P4
P4 R4 P4 R4 P4 | R4 P4 R4 P4
R4             | R4 P4 R4 P4
P4 P0 P0 P0 P0 | P0 P0 P0 P0

(The tokens are not on the same line, but you can check they're the same).

This gives us hope we might be able to write a DEFLATE quine. But we're not close to done yet, since we have to deal with actual file formats and not made up tokens. Read on!

Part II: Zlib and DEFLATE

Zlib usually appends a 2 byte header and a 4 byte checksum to everything it compresses. The 4 byte checksum would make the creation of a quine much more difficult. But luckily, Bubblegum is designed using to utilize the -zlib.MAX_WBITS flag, which skips the header and the checksum! So we just have a raw DEFLATE stream. How does DEFLATE work? The full thing can be a bit complicated, but luckily we only need to pull out the bits that allow us to have our Pn and Rn building blocks.

Part III: The Pn building block

A deflate stream is made up of a series of blocks. Each block starts with the following:

  • BFINAL: 1 bit, set to 1 if it's the last block.
  • BTYPE: 2 bits. All we need to know is that it's 00 for 'no compression' (ie Pn) and 01 for 'fixed compression' (which turns out to map to Rn).

If we have a 'no compression' block, the rest of the bits in the current byte are set to zero and the next bytes look like:

|  LEN  | NLEN  |... LEN bytes of literal data...|

Where LEN is a 2-byte little endian unsigned number of bytes in the literal data, NLEN is the complement of LEN (also unsigned little endian) and we then have N literal bytes. Keeping in mind the first byte is packed from LSB to MSB, this means we can encode the following:

P0 = 00 00 00 ff ff
00000 00 0 | 00000000 | 00000000 | 11111111 | 11111111
^     ^  ^   ^^^^^^^^^^^^^^^^^^^   ^^^^^^^^^^^^^^^^^^^
|     |  |   LEN = 0x0000          NLEN = ~LEN = 0xFFFF
|     |  |
|     |  \- BFINAL = 0 (not final block)
|     \---- BTYPE = 00 (no compression)
\---------- 5 bits padding in block
P4 = 00 14 00 eb ff
00000 00 0 | 00010100 | 00000000 | 11101011 | 11111111
^     ^  ^   ^^^^^^^^^^^^^^^^^^^   ^^^^^^^^^^^^^^^^^^^
|     |  |   LEN = 0x0014          NLEN = ~LEN = 0xFFEB
|     |  |
|     |  \- BFINAL = 0 (not final block)
|     \---- BTYPE = 00 (no compression)
\---------- 5 bits padding in block

Why is P4 printing 0x14 = 20 bytes, you ask, instead of 4? Well, the previous token 'quine' had the units of 1 byte ~ 1 token, but we don't have that luxury. So instead, we have a fixed length of 5 bytes per token, since this is the minimum size of a print token. So 4 tokens is 20 bytes.

Part IV: The Rn building block

The BTYPE = 01 allows us to make queries of the form REPEAT(n, q):

Starting from q bytes away in the output, print n bytes.

It shouldn't be hard to see that REPEAT(n, n) gives us Rn. But there's a problem, since it turns out that R4 = REPEAT(20, 20) only takes up 3 bytes instead of 5! Since we are assuming all our tokens take up 5 bytes for our quine to work, this is no good. However, we can introduce some redundancy - it turns out if we define R4 = REPEAT(10, 20), REPEAT(10, 20), then we do the same thing but now the instruction takes up 5 bytes total!

The way these blocks are actually encoded as bytes is a little complex. I'll annotate the block, and to fill in the gaps read the RFC. For compressed blocks, the data is turned from bits into bytes LSB to MSB with a couple of exceptions.

P4 = 42 88 21 c4 00
01000 01 0 | 1 00010 00 | 001000 01 | 11 00010 0 | 0000000 0
^     ^  ^   ^  ^    ^    ^      ^    ^   ^    ^   ^       ^
|     |  |  [5] |    |    |      |    [8] |    |   padding [8]
[3]   |  \- [1] [4] [3]  [6]    [5]      [7]  [6]
      \---- [2]

[1]: BFINAL: 0 (not end block)
[2]: BTYPE: 01 (fixed compression) 
[3]: Literal code 264 (print 10 bytes...)
[4]: Distance code 8 (starting from 17 + ... )
[5]: Extra distance code bits ( ... 3 bytes back) (= 20 total)
[6]: Literal code 264 (print 10 bytes...)
[7]: Distance code 8 (starting from 17 + ... )
[8]: Extra distance code bits ( ... 3 bytes back) (= 20 total)

So we've got all our building blocks! P0, P4, R4 right? Are we done?

Part V: The final tweak

Well, not so fast. Remember we had a bit saying which block was the end block? It turns out, for Python at least, that we need to include this on the last block, else it messes up our program. And unfortunately, if we let P*0 be a P0 end block token, the following is NOT a quine:

Input           | Output
P0              | 
P0              |
P0              | 
P4 P0 P0 P0 P4  | P0 P0 P0 P4
R4              | P0 P0 P0 P4
P4 R4 P4 R4 P4  | R4 P4 R4 P4
R4              | R4 P4 R4 P4 <-\
P*4 P0 P0 P0 P0 | P0 P0 P0 P0   |
^                               |
          Not the same!

However, if we introduce an R*1, we can fix this quite easily:

Input            | Output
P0               | 
P0               |
P0               | 
P4 P0 P0 P0 P4   | P0 P0 P0 P4
R4               | P0 P0 P0 P4
P4 R4 P4 R4 P4   | R4 P4 R4 P4
R4               | R4 P4 R4 P4
P4 P0 P0 P0 R*1  | P0 P0 P0 R*1
R*1              | R*1

It turns out we can encode R*1 = 03 13 00 00 00, so we are done. Use the following Python program to assemble and verify our DEFLATE quine:

import zlib

P0 = b'\x00\x00\x00\xff\xff'
P4 = b'\x00\x14\x00\xeb\xff'
R4 = b'B\x88!\xc4\x00'
R1_F = b'\x03\x13\x00\x00\x00'

comp = b''
comp += P0
comp += P0
comp += P0
comp += P4 + P0 + P0 + P0 + P4
comp += R4
comp += P4 + R4 + P4 + R4 + P4
comp += R4
comp += P4 + P0 + P0 + P0 + R1_F
comp += R1_F

print(zlib.decompress(comp, -zlib.MAX_WBITS) == comp)

Well done! You are now a certified deflate quine expert™.

  • \$\begingroup\$ Nice write-up. Did you mean P4 instead of P4 in the final fixed version of the quine? \$\endgroup\$
    – user41805
    Sep 1, 2020 at 7:18

√ å ı ¥ ® Ï Ø ¿ , 9 (possibly 11) bytes

79 87  OW

Notice the double space between the 87 and the OW. This is necessary because of the way √ å ı ¥ ® Ï Ø ¿ outputs.

The O command outputs the whole of the stack as numbers

The W command outputs the whole stack as Unicode interpretations of the numbers

The 11 byte solution

The above code will output

===== OUTPUT =====

79 87  OW


-----Program Execution Information-----

Code        : 79 87  OW
Inputs      : []
Stack       : (79,87)
G-Variable  : None
Byte Length : 9
Exit Status : 0
Error       : None


This is obviously not the code inputted but is outputted automatically by the interpreter. If this is disallowed, there is an 11 byte solution that only outputs the required output:

ł 79 87  OW

This will only output

ł 79 87  OW

I'm not sure if the 9 byte answer is acceptable, could someone please tell me in the comments?

  • 2
    \$\begingroup\$ This is a much less trivial quine than usual - nice! \$\endgroup\$
    – isaacg
    Mar 20, 2017 at 4:46
  • \$\begingroup\$ That looks valid (9 byte). I mean the other stuff is just interpreter items that are always there \$\endgroup\$ May 21, 2017 at 12:29
  • \$\begingroup\$ this isn't non-competing because this is a catalogue and any language is fine \$\endgroup\$ May 21, 2017 at 23:17
  • \$\begingroup\$ Gesundheit! Wait... that's a language name? You didn't just sneeze bytes? How do you say that language name in a conversation haha! \$\endgroup\$ Sep 6, 2018 at 16:18

Java 6 - 138 110 106

Since the question says "golf you a quine", I took Steve P's quine and golfed it:

enum Q{X;{String s="enum Q{X;{String s=%c%s%1$c;System.out.printf(s,34,s);}}";System.out.printf(s,34,s);}}

With credits to Trixie Wolf and Volune.
Note: you need to ignore stderr (e.g. 2>/dev/null)

For great good (and justice)!

  • \$\begingroup\$ I can't get this to work. Did you actually try to compile it? I think you need a System.exit() gimmick or it will fail to run properly. I'll add an answer here with my implementation later if I don't hear back from you soon. \$\endgroup\$ Aug 16, 2014 at 4:00
  • \$\begingroup\$ Actually: given the "ignore stderr" comment obv. you did get it to work. I'm very curious how, though. \$\endgroup\$ Aug 16, 2014 at 4:19
  • \$\begingroup\$ @TrixieWolf It works fine here, there is absolutely no compile error. Did you think I would post it without trying it first? :p Anyway, you can only run it with java 6 (or 5), newer versions check for the main method first. \$\endgroup\$ Aug 16, 2014 at 7:52
  • \$\begingroup\$ I'd like to suggest this improvement: enum Q{X;{String s="enum Q{X;{String s=%c%s%1$c;System.out.printf(s,34,s);}}";System.out.printf(s,34,s);}} \$\endgroup\$
    – Volune
    Aug 17, 2014 at 11:47
  • \$\begingroup\$ @aditsu Ah, that makes perfect sense. I'm busy today, but tomorrow I will check to see if mine still functions correctly (I tested it recently but I'll bet it was on J6). It might still work due to the exit() trick. \$\endgroup\$ Aug 17, 2014 at 21:35

MUMPS, 9 bytes

R w $T(R)

This may fall afoul of the "you can't just read the source file and print it" restriction. Let me explain why I say may.

The line of code you see above constitutes a complete MUMPS "routine" (named R), which is sort of like a single source file in a conventional C-like language... but not quite.

The way MUMPS stores its routines is peculiar among programming languages. Routines are not files living in a regular filesystem. Instead, they are data structures internal to the database itself. The line of code I've supplied above is actually stored as part of the MUMPS global named ^ROUTINE (globals are basically trees). The "R" subtree (in MUMPS parlance, "subscript") of that global would look something like this:

^ROUTINE("R",1)="R w $T(R)"

The first entry is the number of lines of code in the routine. The subsequent entries are the lines of code in the routine itself.

Why do I bring this up? Well, this means that in MUMPS, the routines themselves are first-class entries in the database! One can edit routines by directly manipulating the contents of the ^ROUTINE global, just as one can edit any other global. (Indeed, at the most basic level, if your MUMPS environment doesn't come with an editor, you must invent one for yourself that will edit the ^ROUTINE global on your behalf.)

The ability to manipulate routines in MUMPS code is so important that the standard even defines a function whose explicit purpose is to tell you what code is found at a given line of a given routine. That function is named $T[EXT], and if you give it a pointer to a line of code, it will return the code present at that location.

And that's what we do here. We w[rite] the result of a call to $TEXT(R) - that is, the contents of the line at the first line of the routine R - to the output stream, and since R is only one line long, that makes the program a quine.

This program involves no file IO at all. The whole thing is internal to the MUMPS environment. I claim that this is interesting enough to count as a legitimate quine, despite the fact that this has a surface-level resemblance to a program that just reads and prints the source file.


05AB1E, 14 bytes

Shortest proper 05AB1E quine?


With trailing newline.

Try it online!


0              # Push '0'
                   # Stack: ['0']
 "D34çý"       # Push 'D34çý'
                   # Stack: ['0', 'D34çý']
        D      # Duplicate
                   # Stack: ['0', 'D34çý', 'D34çý']
         34ç   # Push '"'
                   # Stack: ['0', 'D34çý', 'D34çý', '"']
            ý  # Join rest of the stack with '"'
                   # Stack: ['0"D34çý"D34çý']
               # Implicit print
  • \$\begingroup\$ Isn't 1 also a proper quine? \$\endgroup\$
    – ovs
    Jan 15, 2017 at 19:24
  • 2
    \$\begingroup\$ @ovs Not by our standard definition. \$\endgroup\$ Jan 25, 2017 at 16:37

Japt, 10 bytes

"iQ ²"iQ ²

Here's how this works:

"iQ ²"      // Take this string.        iQ ²
      iQ    // Insert a quote.          "iQ ²
         ²  // Double.                  "iQ ²"iQ ²
            // Implicitly output.

Test it online!

Of course, any number literal is also a quine because of implicit output.

  • \$\begingroup\$ Does Japt add a newline at the end of implicit output? \$\endgroup\$ Mar 23, 2016 at 4:28
  • \$\begingroup\$ @CalculatorFeline Nope. \$\endgroup\$ Sep 7, 2016 at 2:24

Befunge-93, 15 14 13 bytes


Works in this interpreter. x is an unrecognized command which reflects the instruction pointer.

Thanks to Jo King for saving 1 byte.

This 14 byte version works in FBBI:


Try it online!

  • \$\begingroup\$ This almost works, but doesn't: "gx:#,_:@#/3: (also 13 bytes). \$\endgroup\$
    – jimmy23013
    May 25, 2019 at 4:37

JavaScript REPL, 21 bytes


It technically doesn't read its own file.

… kind of seems like 0 is also a quine for JavaScript the way this is evaluated, though.

  • 6
    \$\begingroup\$ It reads its own source, though. \$\endgroup\$
    – Joey
    May 12, 2011 at 21:15
  • 1
    \$\begingroup\$ ` Uncaught SyntaxError: Unexpected token =>` in Chrome \$\endgroup\$
    – Nakilon
    Jan 14, 2015 at 9:36
  • 7
    \$\begingroup\$ @Nakilon: Use Firefox. \$\endgroup\$
    – Ry-
    Jan 14, 2015 at 16:17
  • 2
    \$\begingroup\$ +1 for the +_+ in the shorter version \$\endgroup\$
    – user48538
    Jan 4, 2016 at 18:31
  • 3
    \$\begingroup\$ Umm... the first one is actually HTML5. \$\endgroup\$ Jun 15, 2016 at 8:07

05AB1E, 13 bytes


Try it online!

Beats all the string-based 05AB1E quines.


2096239            # integer literal
       D           # duplicate
        20B        # convert the copy to base 20, yielding "D20BJ"
           J       # join with the original

Alternative 13 byter:


C, 78 chars

#define Q(S)char*q=#S;S
Q(main(){printf("#define Q(S)char*q=#S;S\nQ(%s)",q);})

This version is shorter than the familiar 79-character C quine and also doesn't assume ASCII. It does still assume that it's safe to not include stdio.h. (Adding an explicit declaration of printf() brings the length up to 103 chars.)


Shell echo-sed quine:

echo sed -eh -es/[\\\(\\\\\\\\\\\)\\\&\\\|]/\\\\\\\\\\\&/g -es/^/echo\\ / -es/$/\\\|/ -eG|
sed -eh -es/[\(\\\\\)\&\|]/\\\\\&/g -es/^/echo\ / -es/$/\|/ -eG

I wanted to write a sed quine, but sed can only work on its input stream, not generate output spontaneously, so this is an echo-sed quine. This 154-character quine uses command-line sed, which automatically makes it hard to read, and uses three different sed commands, as well as two sequences of eleven backslashes in a row. This quine works in bash, ksh, and sh, but not csh or tcsh.


A blatant, and amusing, cheat: echo $BASH_COMMAND

Another, unreasonably silly, cheat: export PROMPT_COMMAND='echo $BASH_COMMAND';$PROMPT_COMMAND


C, 77 chars

Maybe the easiest one in C.


34 is the ASCII decimal for ".

  • \$\begingroup\$ I count 76 bytes. \$\endgroup\$
    – Lynn
    Jan 18, 2017 at 15:06
  • \$\begingroup\$ @Lynn He must have used wc and forgot to exclude the trailing newline :P \$\endgroup\$
    – MD XF
    May 26, 2017 at 16:32

T-SQL 24

This statment reproduces itself in the EVENTINFO column of the output:

dbcc inputbuffer(@@spid)


  • dbcc inputbuffer() - Displays the last statement sent from the client with the specified process id to the current instance of Microsoft SQL Server
  • @@spid - Retrieves the current process id

tested with SQL Server 2008 R2 and 2012; probably working with other versions as well

Online demo: http://www.sqlfiddle.com/#!3/d41d8/2230


QBasic, 76 (110) 54 (72)

Tested with QB64 on Windows 7, with auto-formatting turned off.

READ a$:?a$;:WRITE a$:DATA"READ a$:?a$;:WRITE a$:DATA"

: is a statement separator, and ? is a shortcut for PRINT. The main trick here is using DATA and READ so we don't have to split the string up to add the quotes. Edit: I learned this week about the WRITE command, which outputs strings wrapped in double-quotes--a significant byte-saver here!

Since actual QBasic doesn't let you turn off auto-formatting, here's the same thing with proper formatting in 72 bytes:

READ x$: PRINT x$;: WRITE x$: DATA "READ x$: PRINT x$;: WRITE x$: DATA "

Original versions (76 bytes golfed, 110 formatted):

READ a$:q$=CHR$(34):?a$+q$+a$+q$:DATA"READ a$:q$=CHR$(34):?a$+q$+a$+q$:DATA"


READ a$: q$ = CHR$(34): PRINT a$ + q$ + a$ + q$: DATA "READ a$: q$ = CHR$(34): PRINT a$ + q$ + a$ + q$: DATA "
  • 1
    \$\begingroup\$ Note that this doesn't work with QBasic 1.1 for MS-DOS 6.2: the autoformatter can't be turned off. \$\endgroup\$
    – Mark
    Feb 27, 2015 at 7:58
  • \$\begingroup\$ @Mark Good point. I added a formatted version. \$\endgroup\$
    – DLosc
    Feb 28, 2015 at 22:21
  • 1
    \$\begingroup\$ You can just load the non-formatted file directly though, right? This seems like a limitation of the editor rather than the language itself. \$\endgroup\$
    – 12Me21
    Apr 2, 2018 at 14:54

RProgN, 3 bytes


Try it online!

This exploits a potential flaw in our definition of proper quine:

It must be possible to identify a section of the program which encodes a different part of the program. ("Different" meaning that the two parts appear in different positions.)

Furthermore, a quine must not access its own source, directly or indirectly.

The stack of RProgN is printed backwards, so the first 0 encodes the second 0, and vice versa.

This can be verified empirically; the program




Try it online!

  • 2
    \$\begingroup\$ Oh my, it's actually getting usage. I feel like a proud father. \$\endgroup\$
    – ATaco
    Dec 16, 2016 at 5:26

Klein, 11 + 6 = 17 bytes

3 additional bytes for the topology argument 001 and another 3 for ASCII output -A.

:?/:2+@> "

Try it online!

Let's start with the topology. The 1 at the end indicates that the north and south edges of the code are mapped to each other in reverse. So if the IP leaves the code through the south edge in the leftmost column, it will re-enter through the north edge in the rightmost column. We use this to skip to the end of the program.

:             Duplicate the top of the stack (implicitly zero).
?             Skip the next command if that value is non-zero (which it isn't).
/             Reflect the IP north.
              The IP leaves through the north edge in the third column from
              the left, so it will re-enter from the south edge in the third
              column from the right.
>             Move east.
":?/:2+@> "   Push the code points of the program, except for the quote itself
              to the stack.
:             Duplicate the top of the stack, now a 32 (the space).
?             Skip the next command (the /).
:             Duplicate the top of the stack again.
2+            Add 2, to turn the space into a quote.
@             Terminate the program.

Charcoal, 64 31 32 (because of newlines)

My first answer in charcoal ever!

Similar to /// and other languages, just straight up ascii would print itself. however that is not payload and also boring, so here is an actual quine.

taking a golfing tip from Ascii-only, and my realisation that the second looping is pointless, I have reduced by >50%


Try it online!


(thanks to ascii-only for making most of this.)

A                     α            Assign to a
 ´α´´´A´F´α´⁺´´´´´ι´α             "α´AFα⁺´´ια", but with ´ escape character with each
                                    these are the variable being assigned to, and the
                                    rest of the program that is not the string.

                        ´A         Print A to the grid. current grid: "A"
                           Fα⁺´´ι  For each character in a, print ´ + character
                                    this results in the escaped version of the string
                                    which is the literal string that is assigned at the 
                                    start. current grid state: "A´α´´´A´F´α´⁺´´´´´ι´α"

                                  α Print a ("α´AFα⁺´´ια"), which is the commands after
                                    the string assignment. final grid state vvv:

[implicitly print the grid: "A´α´´´A´F´α´⁺´´´´´ι´αα´AFα⁺´´ια", the source, with a trailing newline]
  • \$\begingroup\$ Wish I was better at reading Charcoal. Looking forward to that explanation :) \$\endgroup\$
    – Emigna
    May 19, 2017 at 10:11
  • 1
    \$\begingroup\$ I can hardly read this myself :P \$\endgroup\$ May 19, 2017 at 10:12
  • \$\begingroup\$ You can leave off the final closing double angle bracket, saving 3 bytes: A´α´´´A´F´L´α´«´´´´´§´α´ι´»´F´L´α´«´§´α´ια´AFLα«´´§αι»FL᫧αι \$\endgroup\$
    – ASCII-only
    May 19, 2017 at 10:33
  • \$\begingroup\$ Oh wait you can also iterate over the string directly. 37 bytes: A´α´´´A´F´α´⁺´´´´´ι´F´α´ια´AFα⁺´´ιFαι \$\endgroup\$
    – ASCII-only
    May 19, 2017 at 10:38
  • \$\begingroup\$ @ASCII-only couldn't this be one byte? f \$\endgroup\$ May 19, 2017 at 14:11

///, 204 bytes


Try it online!

With some helpful whitespace inserted:


How it works

  • The long third line is the quining data. It is made from the entire rest of the program, with a P2 in the spot where the data itself would fit, and then with the string <> inserted before each character from the set \/12.
    • It would be harmless to put <> before all characters in the data, but only these are necessary - \/ because they need escaping to be copied, and 12 because it's vital to have a break inside P1 and P2 to prevent infinite loops when substituting them.
  • The first substitution changes all the <> prefixes into <\\>\\\. The \ in the source <\> is there to prevent its final printable form from being garbled by the other substitutions.
  • The second substitution includes the quining data, copying them to the other P1s in the program. The <\\>\\\ prefixes now become <\>\ in both copies.
  • The third substitution copies one of the quining data copies (in the substitution itself) into the middle of the other (at the end of the program), marked by the string P<\>\2. In the inner copy, the <\>\ prefix now becomes <> again.
  • The fourth substitution changes the inner copy's <> prefixes into <<\>>\. The change is needed to introduce the final backspace, protecting any following \s and /s that are to be printed. The inner <\> is necessary to prevent this substitution from infinitely looping – just a backslash here wouldn't do, as it would be garbled by the fifth substitution.
  • The fifth substitution removes all instances of the string <\>, both those remaining in the outer copy of the quining data, and those produced by the fourth substitution.
  • Finally, we reach the constructed copy of the program, with suitable backslashes prepended to some characters, ready for printing.

Reflections, 4222 bytes

Since wastl out-golfed me by about... 1.810371 bytes through a vastly superior encoding system, I've decided to have another look at the problem. Since my program is still quite long, here's the main section (with SOHs replaced with spaces):

      /# 0v\/(1v/
      \+#@~ > ~<

Try It Online! (but have patience) (ASCII-only points out that unchecking the time between steps will make it go faster, but beware of the javascript freezing up your browser)

This uses the same encoding as wastl's answer, where each character with byte value n is represented by n newlines followed by


and the first character of the code is \ to change the pointer's direction down. Additionally, it also encodes the \ as well as the #,+ and newline in this process to save on doing them later

The main code is a more streamlined version of wastl's, where quite a few shortcuts have been made. I've also replaced all the spaces with SOHs (byte value 1) to save on bytes.

Detailed explanation

\0=0        Create a copy of the data in stack 0
    #_      Print the `\`
      (4:(2(4(4   Push the +, \n, # to stack 4, and a copy of the newline to stack 2
               0\ Switch back to the intact copy of the data

            /(1v/ Reverse the data
            > ~<

          ~   While the stack exists

           1  Move data to stack 1

         4=/  Copy #, \n, +
    (2:2)     Copy newline
\#1)          Get top of data

\#        Redefine origin and move up

+     Push -2

/#@   Print the newline the value of the top of data times
   #_#_#_   Print the +, \n, #
         1^ Switch back to the data and loop again

      /# 0v When the data stack is empty

         0  Switch to the other copy of the data

      /#    Redefine the origin to push 1
        #@  Print the whole stack
          ~ >  And end
  • \$\begingroup\$ You should also say that unchecking the time between steps box shortens run time to like <5 seconds \$\endgroup\$
    – ASCII-only
    May 22, 2018 at 0:58
  • \$\begingroup\$ @ASCII-only Ehh, depends on the computer I guess. Mine freezes up and finishes in about 40 seconds \$\endgroup\$
    – Jo King
    May 22, 2018 at 1:51

J (REPL) - 20 (16?) char

Seems we're missing a J entry. Trivially, any sentence that doesn't evaluate gets itself printed in the REPL, so 1 or + or +/ % # are all quines in that sense. A non-trivial quine would be one that produces specifically a string containing the source code.


u:39 is the ASCII character 39, i.e. the single quote, and ',~@,~u:39' is a string. , is the append verb. The main verb ,~@,~ evaluates as follows:

x ,~@,~ y      
y ,~@, x       NB. x f~ y => y f x       "Passive"
,~ (y , x)     NB. x f@g y => f (x g y)  "At"
(y,x) , (y,x)  NB. f~ y => y f y         "Reflex"

So the result is 'string'string when x is string and y is the single quote, and thus this is a quine when x is ,~@,~u:39.

If we're allowed the J standard library as well, then we can write the 16 character


which appends the quote of the string (,quote) to itself.


APL, 22 bytes


This is part of the FinnAPL Idiom Library.

        '''1⌽22⍴11⍴'''  ⍝ The string literal '1⌽22⍴11⍴' (quotes in string)
     11⍴                ⍝ Fill an 11-element array with these characters
                        ⍝ But the string has length 10, so we get '1⌽22⍴11⍴''
  22⍴                   ⍝ Do this again for 22 chars: '1⌽22⍴11⍴'''1⌽22⍴11⍴''
1⌽                      ⍝ Rotate left (puts quote at the back)

Try it on ngn/apl


Ruby, 26 bytes

printf x="printf x=%p,x",x

Not the shortest Ruby quine, but I think it's quite cute so I'm posting it.

Ruby, 27 bytes

p S if$><<S="p S if$><<S="

Ruby, 28 bytes

printf *["printf *[%p]*2"]*2

Ruby 2.7, 34 bytes

".tap{puts p _1}"
.tap{puts p _1}

Ruby, 35 bytes



TypeScript Types, 149 bytes

type Q<A='`',B='$',C=Q<'${A}','${B}','${C}'>>=`//@ts-nocheck
type Q<A='${A}',B='${B}',C=Q<'${B}{A}','${B}{B}','${B}{C}'>>=${A}${C}${A}`

Try it online!

TypeScript Types, 209 197 bytes

Without @ts-nocheck to suppress compiler errors

type Q<X extends string[]=['`','$',Q<['${X[0]}','${X[1]}','${X[2]}']>]>=`type Q<X extends string[]=['${X[0]}','${X[1]}',Q<['${X[1]}{X[0]}','${X[1]}{X[1]}','${X[1]}{X[2]}']>]>=${X[0]}${X[2]}${X[0]}`

Try it online!


Befunge 98 - 17 11 characters


Or if using g is allowed:

Befunge 98 - 12 10

  • \$\begingroup\$ Explanation for why using g may not be allowed? \$\endgroup\$
    – MD XF
    Jun 9, 2017 at 23:56
  • \$\begingroup\$ @MDXF g is arguably reading the source code. Befunge copies the code to the execution space and g reads the character at the x, y position in the execution space \$\endgroup\$
    – Justin
    Jun 11, 2017 at 6:50

TECO, 20 bytes


The <Esc> should be replaced with ASCII 0x1B, and the <Tab> with 0x09.

  • <Tab>V27:^TJDV<Esc> inserts the text <Tab>V27:^TJDV. This is not because there is a text insertion mode which TECO starts in by default. Instead, <Tab> text <Esc> is a special insertion command which inserts a tab, and then the text. A string whose own initial delimiter is part of the text -- very handy.
  • V prints the current line.
  • 27:^T prints the character with ASCII code 27 without the usual conversion to a printable representation.
  • J jumps to the beginning of the text.
  • D deletes the first character (the tab).
  • V prints the line again.

Lua, 44 bytes


Some other comical answers in Lua:


...so long as the file is named print(arg[0]) And...

Lua: quine.lua:1: function arguments expected near '.'

...so long as the file is named quine.lua


Python 2, 31 bytes

s="print's=%r;exec s'%s";exec s

It's 2 bytes longer than the shortest Python quine on this question, but it's much more useful, since you don't need to write everything twice.

For example, to print a program's own source code in sorted order, we can just do:

s="print''.join(sorted('s=%r;exec s'%s))";exec s

Another example by @feersum can be found here.


The reason the quine works is because of %r's behaviour. With normal strings, %r puts single quotes around the string, e.g.

>>> print "%r"%"abc"

But if you have a single quotes inside the string, it uses double quotes instead:

>>> print "%r"%"'abc'"

This does, however, mean that the quine has a bit of a problem if you want to use both types of quotes in the string.

  • 1
    \$\begingroup\$ Python appends a leading newline to the output, so you will have to add a newline at the end of your program (making your program 32 bytes, not 31). \$\endgroup\$
    – MilkyWay90
    May 4, 2019 at 17:30

RETURN, 18 bytes


Try it here.

First RETURN program on PPCG ever! RETURN is a language that tries to improve DUP by using nested stacks.


"34¤¤,,,,"         Push this string to the stack
          34       Push charcode of " to the stack
            ¤¤     Duplicate top 2 items
              ,,,, Output all 4 stack items from top to bottom

Factor - 74 69 65 bytes

Works on the listener (REPL):

USE: formatting [ "USE: formatting %u dup call" printf ] dup call

This is my first ever quine, I'm sure there must be a shorter one! Already shorter. Now I'm no longer sure... (bad pun attempt)

What it does is:

  • USE: formatting import the formatting vocabulary to use printf
  • [ "U... printf ] create a quotation (or lambda, or block) on the top of the stack
  • dup call duplicate it, and call it

The quotation takes the top of the stack and embeds it into the string as a literal.

Thanks, cat! -> shaved 2 4 more bytes :D

  • 1
    \$\begingroup\$ Welcome to the site. This is a really good answer; however most people replace their old code with the new code and use the edit history to see the old code. You have, however, included a code breakdown and explanation, which not many people do on their first answer, so for that: +1. \$\endgroup\$
    – wizzwizz4
    Feb 14, 2016 at 9:03
  • \$\begingroup\$ @wizzwizz4 Thanks for the advice and up! Actually my 2nd answer, but first quine ever and first edit on PCG. \$\endgroup\$
    – fede s.
    Feb 14, 2016 at 22:07
  • \$\begingroup\$ Well, if you ever need help, feel free to ping me. \$\endgroup\$
    – wizzwizz4
    Feb 14, 2016 at 22:11
  • \$\begingroup\$ I never realised a quine was so simple in Factor! Also, the bottom, shorter one can be a single line for 65 bytes, because you don't need the trailing newline: USE: formatting [ "USE: formatting %u dup call" printf ] dup call \$\endgroup\$
    – cat
    May 17, 2016 at 22:55
  • \$\begingroup\$ Thanks, @cat Just assumed it expected EOL, but this makes more sense actually! \$\endgroup\$
    – fede s.
    May 17, 2016 at 23:07

F#, 90 bytes

let q="let q=%A

F#’s smart printf comes back to byte us! We can’t write let q="...";;printf q q, as the first parameter to printf isn’t actually a string:

printf : TextWriterFormat<'T> -> 'T

F# uses some compiler magic under the hood to guarantee type-safe printf calls. For example, "yay %d wow!" is a valid TextWriterFormat<int -> unit> literal, but not a valid TextWriterFormat<double -> unit> literal. But if we define the format string separately, the compiler will see it as a regular old string and complain. Instead, we have to convert q ourselves in the first argument.

What about let q:TextWriterFormat<_>="..."? First of all, that’s two bytes longer. But second of all, the second argument to printf really needs to be a string, otherwise the typechecker will infer that we’re formatting a formatter, which in turn formats a formatter, which formats a…

error FS0001: Type mismatch. Expecting a
but given a
    Printf.TextWriterFormat<('a -> unit)>    
The resulting type would be infinite when unifying ''a' and
    'Printf.TextWriterFormat<('a -> unit)>'

Yep, an infinite type. Oops.

  • \$\begingroup\$ +1 for emoticon in the code <_> \$\endgroup\$
    – user48538
    Aug 3, 2016 at 18:32
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