222
\$\begingroup\$

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 QUESTION_ID=69;
var OVERRIDE_USER=98;

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)
if(languages.hasOwnProperty(lang))
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> 

\$\endgroup\$
3
  • 4
    \$\begingroup\$ Do you not mean, "Golf you a quine for greater good!"? \$\endgroup\$ May 3 '11 at 2:49
  • 58
    \$\begingroup\$ @muntoo it's a play on "Learn you a Haskell for Great Good". \$\endgroup\$ May 3 '11 at 2:52
  • 10
    \$\begingroup\$ Did anybody notice that this is question 69? \$\endgroup\$
    – aidan0626
    Oct 24 '20 at 22:47

402 Answers 402

1 2
3
4 5
14
10
\$\begingroup\$

√ 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?

\$\endgroup\$
4
  • 2
    \$\begingroup\$ This is a much less trivial quine than usual - nice! \$\endgroup\$
    – isaacg
    Mar 20 '17 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 '17 at 12:29
  • \$\begingroup\$ this isn't non-competing because this is a catalogue and any language is fine \$\endgroup\$ May 21 '17 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 '18 at 16:18
9
\$\begingroup\$

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)!

\$\endgroup\$
6
  • \$\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 '14 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 '14 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 '14 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 '14 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 '14 at 21:35
9
\$\begingroup\$

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",0)=1
^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.

\$\endgroup\$
9
+200
\$\begingroup\$

05AB1E, 14 bytes

Shortest proper 05AB1E quine?

0"D34çý"D34çý

With trailing newline.

Try it online!

Explanation:

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
\$\endgroup\$
2
9
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ Does Japt add a newline at the end of implicit output? \$\endgroup\$ Mar 23 '16 at 4:28
  • \$\begingroup\$ @CalculatorFeline Nope. \$\endgroup\$ Sep 7 '16 at 2:24
9
\$\begingroup\$

Mini-Flak, 6900 bytes

Mini-flak is a Turing complete subset of Brain-Flak. It works exactly like Brain-Flak except the [], <> and <...> operations are banned from use. Programming in Min-Flak is thus much more difficult than traditional Brain-Flak.

The main difficulty with Mini-Flak is the lack of random access. While Mini-flak is Turing complete, location of access (relative to the top of the stack) must be determined at compile time rather than run time.


The following is the quine. Unfortunately this quine has an order notation of O(7**n) (where n is its own length) and thus cannot be run to completion in the lifetime of the universe. I will hopefully convince you that it does work but for now you will have to trust me a bit. If you want a version that can be run in the lifetime of the universe (or an afternoon) you can scroll down a bit to my faster version.

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Explanation

Like my previous Brain-Flak quine This program has two parts. The first part pushes numbers between 1 and 6 to the stack representing the second part of the program using the following key.

1 -> (
2 -> )
3 -> [
4 -> ]
5 -> {
6 -> }

(Since there is no <> in Mini-Flak those characters are left unencoded). It does this in a deterministic fashion so that this section can be reversed by the next section.

The second section is a decoder. It takes the output from the first section and turns it into the code that generates that list and the code represented by that list (this section's source). However this is easier said than done. Because of Mini-Flak's lack of random access we are going to need to abandon Brain-Flak's traditional techniques in favor of some more bizarre methods. This program starts by compressing the entire stack into one base 7 number where each digit is one number in the list. It does that with the following code:

(({}({}))[({}[{}])]){(({}({}))[({}[{}])])((((({})){})){}{}{}{})(({}({}))[({}[{}])])}{}

Try it Online!

This is a pretty straightforward (as far as Mini-Flak goes) program and I won't get into how it works unless any one is interested. (It is a neat little program but to save space I will leave it out).

We now have one single number representing the entire program. I will push a copy to "temporary memory" (the scope) like follows:

(({})[(...)]{})

And decompose the original copy via repeated devision. Each time I remove a digit from the number I will convert it to the code that generates it.

Once I am done with that, the program will put the copy stored in temporary memory back down and begin a second decomposition. This time it will map each digit to the ASCII value of its corresponding brace as it is decomposed from the total.

Once that is done the program has constructed it's source so it simply terminates.


Verification

You might be suspicious of my program. How can we know that it actually works if it won't terminate in the lifetime of the universe?

So I have set up a "toy version" of the original quine to demonstrate that all of the parts are working.

Try it Online!

This version has the first part removed. You can pass the list of numbers that would be generated by the first part as command line arguments. It will construct code that pushes them and the code they represent. I provided a simple test case but I encourage you to try it out with your own! You will notice even with only six characters the run times are starting to become noticeably long. This is because the division I use is O(n). Slow division has always been a reality in Brain-Flak and it carries over into Mini-Flak.

If you have any questions or confusions comment them and I will be happy to address them.


106656 bytes

Now for my fast version.

This version takes about half an hour (175300470 Brain-Flak cycles) to run on my machine using the ruby interpreter. But for the best performance I suggest you use Crain-Flak the C interpreter which is much faster but lacks some of the polish of the ruby interpreter.

Try it online

Explanation

The reason that Miniflak quines are destined to be slow is Miniflak's lack of random access. In the short but slow version (short is a bit of an exaggeration and slow an understatement) I get around this by pushing all the numbers and then packaging them up into one number and unrolling it piece by piece. However this version does it quite differently. I create a block of code that takes in a number and returns a datum. Each datum represents a single character like before and the main code simply queries this block for each one at a time. This essentially works as a block of random access memory.


To construct this block I actually reused a method from my proof that Miniflak is Turing complete. For each datum there is a block of code that looks like this:

(({}[()])[(())]()){(([({}{})]{}))}{}{(([({}{}(%s))]{}))}{}

This subtracts one from the number on top of the stack and if zero pushes %s the datum beneath it. Since each piece decrements the size by one if you start with n on the stack you will get back the nth datum.

This is nice and modular, so it can be written by a program easily.


Next we have to set up the machine that actually translates this memory into the source. This consists of 5 parts as such:

([()]())(()()()())
{({}[(
   -
 )]{})
 1. (({}[()])[((()))]{}){(([({}{})]{}))}{}{([({}{}{}{}
     (((((((((((((((((((((((((((()()()()()){}){}){})((((()()()()){}){}())){}{})(((()()()()){}){}()){})[()()])[((((()()()()()){}){}){}()){}()])((((()()()()()){}){}){}()){}())[((((()()()()()){}){}){}()){}]))[()])())[()])())[()])())[()]))()))[()])((((()()()){}){}()){}){}())[((((()()()){}){}()){}){}])[()])((((()()()()){}){}())){}{})[((((()()()()){}){}())){}{}])
     (()()()())
    )]{})}{}
 2. (({}[()])[((()))]{}){(([({}{})]{}))}{}{([({}{}
     (({}(({}({}))[({}[{}])][(
     ({}[()(
      ([()](((()()[(((((((()()()){})())){}{}){}){})]((((()()()()())){}{}){})([{}]([()()](({})(([{}](()()([()()](((((({}){}){}())){}){}{}))))))))))))
     )]{})
     {({}[()(((({})())[()]))]{})}{}
     (([(((((()()()()){}){}()))){}{}([({})]((({})){}{}))]()()([()()]({}(({})([()]([({}())](({})([({}[()])]()(({})(([()](([({}()())]()({}([()](([((((((()()()())()){}){}){}()){})]({}()(([(((((({})){}){}())){}{})]({}([((((({}())){}){}){}()){}()](([()()])(()()({}(((((({}())())){}{}){}){}([((((({}))){}()){}){}]([((({}[()])){}{}){}]([()()](((((({}())){}{}){}){})(([{}](()()([()()](()()(((((()()()()()){}){}){}()){}()(([((((((()()()())){}){}())){}{})]({}([((((({})()){}){}){}()){}()](([()()])(()()({}(((((({}){}){}())){}){}{}(({})))))))))))))))))))))))))))))))))))))))))))))))
     )]{})[()]))({()([({})]{})}{}()()()())
    )]{})}{}
 3. (({}[()])[((()))]{}){(([({}{})]{}))}{}{([({}{}
      (({}[(
      ({}[()(((((()()()()()){}){}){}))]{}){({}[()(({}()))]{}){({}[()(({}((((()()()){}){}){}()){}))]{}){({}[()(({}()()))]{}){({}[()(({}(((()()()()())){}{}){}))]{}){([(({}{}()))]{})}}}}}{}
      (({}({}))[({}[{}])])
     )]{}({})[()]))
      ({()([({}({}[({})]))]{})}{}()()()()[(({}({})))]{})
    )]{})}{}
 4. (({}[()])[((()))]{}){(([({}{})]{}))}{}{([({}{}(([{}]))(()()()()))]{})}{}
    ({}[()])
}{}

The machine consists of four parts that are run in reverse starting with 4 and ending with 1. I have labeled them in the code above. Each section also uses the same lookup table format I use for the encoding. This is because the entire program is contained in a loop and we don't want to run every section every time we run through the loop so we put in the same RA structure and query the section we desire each time.

4

Section 4 is a simple set up section.

The program tells first queries section 4 and datum 0. Datum 0 does not exist so instead of returning that value it simply decrements the query once for each datum. This is useful because we can use the result to determine the number of data, which will become important in future sections. Section 4 records the number of data by negativizing the result and queries Section 3 and the last datum. The only problem is we cannot query section 3 directly. Since there is another decrement left we need to query a section 4. In fact this will be the case every time we query a section within another section. I will ignore this in my explanation however if you are looking a the code just remember 4 means go back a section and 5 means run the same section again.

3

Section 3 decodes the data into the characters that make up the code after the data block. Each time it expects the stack to appear as so:

Previous query
Result of query
Number of data
Junk we shouldn't touch...

It maps each possible result (a number from 1 to 6) to one of the six valid Miniflak characters ((){}[]) and places it below the number of data with the "Junk we shouldn't touch". This gets us a stack like:

Previous query
Number of data
Junk we shouldn't touch...

From here we need to either query the next datum or if we have queried them all move to section 2. Previous query is not actually the exact query sent out but rather the query minus the number of data in the block. This is because each datum decrements the query by one so the query comes out quite mangled. To generate the next query we add a copy of the number of data and subtract one. Now our stack looks like:

Next query
Number of data
Junk we shouldn't touch...

If our next query is zero we have read all the memory needed in section 3 so we add the number of data to the query again and slap a 4 on top of the stack to move onto section 2. If the next query is not zero we put a 5 on the stack to run section 3 again.

2

Section 2 makes the block of data by querying our RAM just as section 3 does.

For the sake of brevity I will omit most of the details of how section 2 works. It is almost identical to section 3 except instead of translating each datum into one character it translates each into a lengthy chunk of code representing its entry in the RAM. When section 2 is done it calls on section 1.

1

Section one is the most simple section.

It pushes the first bit of the quine ([()]())(()()()()){({}[( and defers to section 5.

5

There is no real section 5 instead a 5 will be decremented once by each section, entering none of them and the once more by the decrement hanging around at the end of the loop. This will result in a zero and will exit the main loop terminating the program.


I hope this was clear. Please comment if you are confused about anything.

\$\endgroup\$
4
  • 2
    \$\begingroup\$ Wait... so the quine of a subset is shorter than the quine of the original? \$\endgroup\$ Dec 8 '16 at 23:52
  • \$\begingroup\$ @ETHproductions No, the original was golfed down to 11k \$\endgroup\$
    – DJMcMayhem
    Dec 9 '16 at 0:14
  • \$\begingroup\$ @DJMcMayhem But isn't 6900 less than 11028? \$\endgroup\$ Dec 9 '16 at 0:15
  • \$\begingroup\$ ... Apparently I can't math... \$\endgroup\$
    – DJMcMayhem
    Dec 9 '16 at 0:16
9
\$\begingroup\$

Befunge-93, 15 14 13 bytes

+9*5x:#,_:@#"

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:

+9*5<>:#,_:@#"

Try it online!

\$\endgroup\$
1
  • \$\begingroup\$ This almost works, but doesn't: "gx:#,_:@#/3: (also 13 bytes). \$\endgroup\$
    – jimmy23013
    May 25 '19 at 4:37
9
\$\begingroup\$

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.

\$\endgroup\$
6
  • 6
    \$\begingroup\$ It reads its own source, though. \$\endgroup\$
    – Joey
    May 12 '11 at 21:15
  • 1
    \$\begingroup\$ ` Uncaught SyntaxError: Unexpected token =>` in Chrome \$\endgroup\$
    – Nakilon
    Jan 14 '15 at 9:36
  • 7
    \$\begingroup\$ @Nakilon: Use Firefox. \$\endgroup\$
    – Ry-
    Jan 14 '15 at 16:17
  • 2
    \$\begingroup\$ +1 for the +_+ in the shorter version \$\endgroup\$
    – user48538
    Jan 4 '16 at 18:31
  • 3
    \$\begingroup\$ Umm... the first one is actually HTML5. \$\endgroup\$ Jun 15 '16 at 8:07
9
\$\begingroup\$

05AB1E, 13 bytes

2096239D20BJ
    

Try it online!

Beats all the string-based 05AB1E quines.

Explanation:

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

Alternative 13 byter:

11959189D₃BJ
    
\$\endgroup\$
1
8
\$\begingroup\$

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.)

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

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.

EDIT:

A blatant, and amusing, cheat: echo $BASH_COMMAND

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

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

C, 77 chars

Maybe the easiest one in C.

main(){char*c="main(){char*c=%c%s%c;printf(c,34,c,34);}";printf(c,34,c,34);}

34 is the ASCII decimal for ".

\$\endgroup\$
2
  • \$\begingroup\$ I count 76 bytes. \$\endgroup\$
    – Lynn
    Jan 18 '17 at 15:06
  • \$\begingroup\$ @Lynn He must have used wc and forgot to exclude the trailing newline :P \$\endgroup\$
    – MD XF
    May 26 '17 at 16:32
8
\$\begingroup\$

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"

or

READ a$: q$ = CHR$(34): PRINT a$ + q$ + a$ + q$: DATA "READ a$: q$ = CHR$(34): PRINT a$ + q$ + a$ + q$: DATA "
\$\endgroup\$
3
  • 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 '15 at 7:58
  • \$\begingroup\$ @Mark Good point. I added a formatted version. \$\endgroup\$
    – DLosc
    Feb 28 '15 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 '18 at 14:54
8
\$\begingroup\$

RProgN, 3 bytes

0
0

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

1
2

prints

2
1

Try it online!

\$\endgroup\$
1
  • 2
    \$\begingroup\$ Oh my, it's actually getting usage. I feel like a proud father. \$\endgroup\$
    – ATaco
    Dec 16 '16 at 5:26
8
\$\begingroup\$

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.
\$\endgroup\$
8
\$\begingroup\$

///, 204 bytes

/<\>/<\\\\>\\\\\\//P1/<>/<<>\><>/<<>\<>\<>\<>\><>\<>\<>\<>\<>\<>\<>/<>/P<>1<>/P<>2<>/<>/P<<>\<>\><>\<>\<>2<>/P<>1<>/<>/<<>\><>/<<<>\<>\>><>\<>\<>/<>/<<>\<>\><>/<>/P<>1//P<\\>\\2/P1//<\>/<<\\>>\\//<\\>//P1

Try it online!

With some helpful whitespace inserted:

/<\>/<\\\\>\\\\\\/
/P1/
    <>/<<>\><>/<<>\<>\<>\<>\><>\<>\<>\<>\<>\<>\<>/<>/P<>1<>/P<>2<>/<>/P<<>\<>\><>\<>\<>2<>/P<>1<>/<>/<<>\><>/<<<>\<>\>><>\<>\<>/<>/<<>\<>\><>/<>/P<>1
/
/P<\\>\\2/P1/
/<\>/<<\\>>\\/
/<\\>//
P1

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.
\$\endgroup\$
8
+200
\$\begingroup\$

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):

\0=0#_(4:(2(4(40\
      /# 0v\/(1v/
      \+#@~ > ~<
/#@#_#_#_1^1/
+
\#1)(2:2)4=/

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
            > ~<
          ^1/


          v\
          ~   While the stack exists
          ^

          v\
              
           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
      \+#@~
/#@#_#_#_1^

         0  Switch to the other copy of the data

      /#    Redefine the origin to push 1
      \+
        #@  Print the whole stack
          ~ >  And end
\$\endgroup\$
2
  • \$\begingroup\$ You should also say that unchecking the time between steps box shortens run time to like <5 seconds \$\endgroup\$
    – ASCII-only
    May 22 '18 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 '18 at 1:51
8
\$\begingroup\$

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',~@,~u:39

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

(,quote)'(,quote)'

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

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

APL, 22 bytes

1⌽22⍴11⍴'''1⌽22⍴11⍴'''

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

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

TECO, 20 bytes

<Tab>V27:^TJDV<Esc>V27:^TJDV

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.
\$\endgroup\$
1
7
\$\begingroup\$

T-SQL 24

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

dbcc inputbuffer(@@spid)

Explanation:

  • 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

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

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.

Notes

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"
'abc'

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

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

\$\endgroup\$
1
  • 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 '19 at 17:30
7
\$\begingroup\$

RETURN, 18 bytes

"34¤¤,,,,"34¤¤,,,,

Try it here.

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

Explanation

"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
\$\endgroup\$
0
7
\$\begingroup\$

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

\$\endgroup\$
9
  • 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 '16 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 '16 at 22:07
  • \$\begingroup\$ Well, if you ever need help, feel free to ping me. \$\endgroup\$
    – wizzwizz4
    Feb 14 '16 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 '16 at 22:55
  • \$\begingroup\$ Thanks, @cat Just assumed it expected EOL, but this makes more sense actually! \$\endgroup\$
    – fede s.
    May 17 '16 at 23:07
7
\$\begingroup\$

F#, 90 bytes

let q="let q=%A
printf(Printf.TextWriterFormat<_>q)q"
printf(Printf.TextWriterFormat<_>q)q

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
    '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.

\$\endgroup\$
1
  • \$\begingroup\$ +1 for emoticon in the code <_> \$\endgroup\$
    – user48538
    Aug 3 '16 at 18:32
7
\$\begingroup\$

S.I.L.O.S, 2642 2593 bytes

Credits to Rohan Jhunjhunwala for the algorithm.

A = 99
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 76
A + 1
set A 105
A + 1
set A 110
A + 1
set A 101
A + 1
set A 32
A + 1
set A 65
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 57
A + 1
set A 57
A + 1
set A 10
A + 1
set A 67
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 57
A + 1
set A 57
A + 1
set A 10
A + 1
set A 66
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 67
A + 1
set A 10
A + 1
set A 108
A + 1
set A 98
A + 1
set A 108
A + 1
set A 68
A + 1
set A 10
A + 1
set A 67
A + 1
set A 32
A + 1
set A 43
A + 1
set A 32
A + 1
set A 49
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 32
A + 1
set A 115
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 65
A + 1
set A 32
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 73
A + 1
set A 110
A + 1
set A 116
A + 1
set A 32
A + 1
set A 66
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 76
A + 1
set A 105
A + 1
set A 110
A + 1
set A 101
A + 1
set A 32
A + 1
set A 65
A + 1
set A 32
A + 1
set A 43
A + 1
set A 32
A + 1
set A 49
A + 1
set A 10
A + 1
set A 66
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 67
A + 1
set A 10
A + 1
set A 105
A + 1
set A 102
A + 1
set A 32
A + 1
set A 66
A + 1
set A 32
A + 1
set A 68
A + 1
set A 10
A + 1
set A 70
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 57
A + 1
set A 57
A + 1
set A 10
A + 1
set A 69
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 70
A + 1
set A 10
A + 1
set A 108
A + 1
set A 98
A + 1
set A 108
A + 1
set A 71
A + 1
set A 10
A + 1
set A 70
A + 1
set A 32
A + 1
set A 43
A + 1
set A 32
A + 1
set A 49
A + 1
set A 10
A + 1
set A 112
A + 1
set A 114
A + 1
set A 105
A + 1
set A 110
A + 1
set A 116
A + 1
set A 67
A + 1
set A 104
A + 1
set A 97
A + 1
set A 114
A + 1
set A 32
A + 1
set A 69
A + 1
set A 10
A + 1
set A 69
A + 1
set A 32
A + 1
set A 61
A + 1
set A 32
A + 1
set A 103
A + 1
set A 101
A + 1
set A 116
A + 1
set A 32
A + 1
set A 70
A + 1
set A 10
A + 1
set A 105
A + 1
set A 102
A + 1
set A 32
A + 1
set A 69
A + 1
set A 32
A + 1
set A 71
A + 1
printLine A = 99
C = 99
B = get C
lblD
C + 1
print set A 
printInt B
printLine A + 1
B = get C
if B D
F = 99
E = get F
lblG
F + 1
printChar E
E = get F
if E G

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ Good job! May I "borrow this post for my github repository? \$\endgroup\$ Aug 26 '16 at 19:12
  • \$\begingroup\$ @RohanJhunjhunwala Sure. \$\endgroup\$
    – Leaky Nun
    Aug 26 '16 at 19:18
7
\$\begingroup\$

S.I.L.O.S, 3057 bytes

A = 99
def S set 
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 76
A + 1
S A 105
A + 1
S A 110
A + 1
S A 101
A + 1
S A 32
A + 1
S A 65
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 57
A + 1
S A 57
A + 1
S A 10
A + 1
S A 72
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 56
A + 1
S A 51
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 32
A + 1
S A 100
A + 1
S A 101
A + 1
S A 102
A + 1
S A 32
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 67
A + 1
S A 104
A + 1
S A 97
A + 1
S A 114
A + 1
S A 32
A + 1
S A 72
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 76
A + 1
S A 105
A + 1
S A 110
A + 1
S A 101
A + 1
S A 32
A + 1
S A 32
A + 1
S A 83
A + 1
S A 32
A + 1
S A 10
A + 1
S A 67
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 57
A + 1
S A 57
A + 1
S A 10
A + 1
S A 66
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 67
A + 1
S A 10
A + 1
S A 108
A + 1
S A 98
A + 1
S A 108
A + 1
S A 68
A + 1
S A 10
A + 1
S A 67
A + 1
S A 32
A + 1
S A 43
A + 1
S A 32
A + 1
S A 49
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 67
A + 1
S A 104
A + 1
S A 97
A + 1
S A 114
A + 1
S A 32
A + 1
S A 72
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 32
A + 1
S A 32
A + 1
S A 65
A + 1
S A 32
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 73
A + 1
S A 110
A + 1
S A 116
A + 1
S A 32
A + 1
S A 66
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 76
A + 1
S A 105
A + 1
S A 110
A + 1
S A 101
A + 1
S A 32
A + 1
S A 65
A + 1
S A 32
A + 1
S A 43
A + 1
S A 32
A + 1
S A 49
A + 1
S A 10
A + 1
S A 66
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 67
A + 1
S A 10
A + 1
S A 105
A + 1
S A 102
A + 1
S A 32
A + 1
S A 66
A + 1
S A 32
A + 1
S A 68
A + 1
S A 10
A + 1
S A 70
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 57
A + 1
S A 57
A + 1
S A 10
A + 1
S A 69
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 70
A + 1
S A 10
A + 1
S A 108
A + 1
S A 98
A + 1
S A 108
A + 1
S A 71
A + 1
S A 10
A + 1
S A 70
A + 1
S A 32
A + 1
S A 43
A + 1
S A 32
A + 1
S A 49
A + 1
S A 10
A + 1
S A 112
A + 1
S A 114
A + 1
S A 105
A + 1
S A 110
A + 1
S A 116
A + 1
S A 67
A + 1
S A 104
A + 1
S A 97
A + 1
S A 114
A + 1
S A 32
A + 1
S A 69
A + 1
S A 10
A + 1
S A 69
A + 1
S A 32
A + 1
S A 61
A + 1
S A 32
A + 1
S A 103
A + 1
S A 101
A + 1
S A 116
A + 1
S A 32
A + 1
S A 70
A + 1
S A 10
A + 1
S A 105
A + 1
S A 102
A + 1
S A 32
A + 1
S A 69
A + 1
S A 32
A + 1
S A 71
A + 1
printLine A = 99
H = 83
print def 
printChar H
printLine  S 
C = 99
B = get C
lblD
C + 1
printChar H
print  A 
printInt B
printLine A + 1
B = get C
if B D
F = 99
E = get F
lblG
F + 1
printChar E
E = get F
if E G

Try it online!

I am ashamed to say this took me a while to write even though most of it was generated by another java program. Thanks to @MartinEnder for helping me out. This is the first quine I have ever written. Credits go to Leaky Nun for most of the code. I "borrowed his code" which was originally inspired by mine. My answer is similar to his, except it shows the "power" of the preprocessor. Hopefully this approach can be used to golf of bytes if done correctly. The goal was to prevent rewriting the word "set" 100's of times.
Please check out his much shorter answer!

\$\endgroup\$
4
  • \$\begingroup\$ How does this work? \$\endgroup\$
    – Leaky Nun
    Aug 26 '16 at 18:44
  • \$\begingroup\$ It's borked from my understanding @LeakyNun but it essentially writes it source code to the memory buffer, and then prints out commands to write itself to the memory buffer, and then writes itself out \$\endgroup\$ Aug 26 '16 at 18:46
  • \$\begingroup\$ From my point of view this is not a quine? \$\endgroup\$
    – Leaky Nun
    Aug 26 '16 at 18:55
  • \$\begingroup\$ @LeakyNun it's borked... let me fix... should I delete, fix and undelete? \$\endgroup\$ Aug 26 '16 at 18:58
7
\$\begingroup\$

ಠ_ಠ, 6 bytes

ಠಠ

This used to work back when the interpreter was still buggy but that's fixed now. However, you can try it in the legacy version of the interpreter!

\$\endgroup\$
3
  • 2
    \$\begingroup\$ It seems that also works in the interpreter provided. Also, ಠ is 3 bytes in UTF-8 \$\endgroup\$ Aug 19 '16 at 21:53
  • \$\begingroup\$ Your link is dead, and can you upload the language source to GitHub so that it can be added to TIO? \$\endgroup\$
    – Pavel
    Jan 30 '17 at 4:28
  • \$\begingroup\$ codepen.io/molarmanful/pen/rxJmqx all the code happens to be there. \$\endgroup\$ Jan 30 '17 at 17:58
7
+100
\$\begingroup\$

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%

A´α´´´A´F´α´⁺´´´´´ι´αα´AFα⁺´´ια

Try it online!

Explanation

(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
                                    character
                                    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:
                                                 "A´α´´´A´F´α´⁺´´´´´ι´αα´AFα⁺´´ια"

[implicitly print the grid: "A´α´´´A´F´α´⁺´´´´´ι´αα´AFα⁺´´ια", the source, with a trailing newline]
\$\endgroup\$
15
  • \$\begingroup\$ Wish I was better at reading Charcoal. Looking forward to that explanation :) \$\endgroup\$
    – Emigna
    May 19 '17 at 10:11
  • 1
    \$\begingroup\$ I can hardly read this myself :P \$\endgroup\$ May 19 '17 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 '17 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 '17 at 10:38
  • \$\begingroup\$ @ASCII-only couldn't this be one byte? f \$\endgroup\$ May 19 '17 at 14:11
7
\$\begingroup\$

Husk, 8 bytes

S+s"S+s"

Try it online!

Husk is a new golfing functional language created by me and Zgarb. It is based on Haskell, but has an intelligent inferencer that can "guess" the intended meaning of functions used in a program based on their possible types.

Explanation

This is a quite simple program, composed by just three functions:

S is the S combinator from SKI (typed) combinator calculus: it takes two functions and a third value as arguments and applies the first function to the value and to the second function applied to that value (in code: S f g x = f x (g x)).

This gives us +"S+s"(s"S+s"). s stands for show, the Haskell function to convert something to a string: if show is applied to a string, special characters in the string are escaped and the whole string is wrapped in quotes.

We get then +"S+s""\"S+s\"". Here, + is string concatenation; it could also be numeric addition, but types wouldn't match so the other meaning is chosen by the inferencer.

Our result is then "S+s\"S+s\"", which is a string that gets printed simply as S+s"S+s".

\$\endgroup\$
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