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Your task is to create the shortest infinite loop!

The point of this challenge is to create an infinite loop producing no output, unlike its possible duplicate. The reason to this is because the code might be shorter if no output is given.

Rules

  • Each submission must be a full program.
  • You must create the shortest infinite loop.
  • Even if your program runs out of memory eventually, it is still accepted as long as it is running the whole time from the start to when it runs out of memory. Also when it runs out of memory, it should still not print anything to STDERR.
  • The program must take no input (however, reading from a file is allowed), and should not print anything to STDOUT. Output to a file is also forbidden.
  • The program must not write anything to STDERR.
  • Feel free to use a language (or language version) even if it's newer than this challenge. -Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language. :D
  • Submissions are scored in bytes, in an appropriate (pre-existing) encoding, usually (but not necessarily) UTF-8. Some languages, like Folders, are a bit tricky to score - if in doubt, please ask on Meta.
  • This is not about finding the language with the shortest infinite loop program. This is about finding the shortest infinite loop program in every language. Therefore, I will not accept an answer.
  • If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainf**k-derivatives like Alphuck), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.
  • There should be a website such as Wikipedia, Esolangs, or GitHub for the language. For example, if the language is CJam, then one could link to the site in the header like #[CJam](http://sourceforge.net/p/cjam/wiki/Home/), X bytes.
  • Standard loopholes are not allowed.

(I have taken some of these rules from Martin Büttner's "Hello World" challenge)


Please feel free to post in the comments to tell me how this challenge could be improved.

Catalogue

This is a Stack Snippet which generates both an alphabetical catalogue of the used languages, and an overall leaderboard. To make sure your answer shows up, please start it with this Markdown header:

# Language name, X bytes

Obviously replacing Language name and X bytes with the proper items. If you want to link to the languages' website, use this template, as posted above:

#[Language name](http://link.to/the/language), X bytes

Now, finally, here's the snippet: (Try pressing "Full page" for a better view.)

var QUESTION_ID=59347;var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk";var OVERRIDE_USER=41805;var answers=[],answers_hash,answer_ids,answer_page=1,more_answers=true,comment_page;function answersUrl(index){return"//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"//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:true,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=false;comment_page=1;getComments()}})}function getComments(){jQuery.ajax({url:commentUrl(comment_page++,answer_ids),method:"get",dataType:"jsonp",crossDomain:true,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=/<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/;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,});else console.log(body)});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('<a>'+lang+'</a>').text();languages[lang]=languages[lang]||{lang:a.language,lang_raw:lang,user:a.user,size:a.size,link:a.link}});var langs=[];for(var lang in languages)if(languages.hasOwnProperty(lang))langs.push(languages[lang]);langs.sort(function(a,b){if(a.lang_raw.toLowerCase()>b.lang_raw.toLowerCase())return 1;if(a.lang_raw.toLowerCase()<b.lang_raw.toLowerCase())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;width:500px;float:left}#language-list{padding:10px;padding-right:40px;width:500px;float:left}table thead{font-weight:700}table td{padding:5px}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution 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>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table>

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  • 52
    \$\begingroup\$ I've got to start posting programs with a negative byte count to beat all these empty files! \$\endgroup\$
    – CJ Dennis
    Oct 3, 2015 at 4:32
  • 4
    \$\begingroup\$ This challenge is interesting because it brings out lots of 0 byte languages (some of which are NOT esolangs). FWIW, most declarative languages have an implicit infinite loop because declarative languages don't have loops in their syntax (they assume they're running in an infinite loop). Ladder diagrams are perhaps among the oldest such languages. Then you have the Instruction Language (IL), a sort of assembly for PLCs that also assume an infinite loop. ILs, like assembly are different between manufacturers \$\endgroup\$
    – slebetman
    Oct 5, 2015 at 9:36
  • \$\begingroup\$ Are programs that read and execute their own source code allowed, or does file I/O break the "must take no input" rule? \$\endgroup\$ Oct 6, 2015 at 13:05
  • 1
    \$\begingroup\$ @ThisSuitIsBlackNot Yes, file input is allowed. \$\endgroup\$
    – user41805
    Oct 6, 2015 at 16:47
  • \$\begingroup\$ Can you print "", an empty string? \$\endgroup\$
    – AAM111
    Mar 7, 2016 at 23:39

568 Answers 568

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1
\$\begingroup\$

vJASS (Warcraft 3), 58 bytes

//! inject main
loop
endloop
//! dovjassinit
//! endinject

Without exitwhen inside the loop, it loops infinitely.

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1
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Pxem, Filename: 4 bytes + Content: 0 bytes = 4 bytes.

  • Filename: .w.a
  • Content: none
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1
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BRASCA, 3 2 bytes

1J

Try it!

Explanation

1J - Move the pointer back one character.
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1
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convey, 2 bytes

5]

Try it online!

In convey, } is an output, and ] is a dummy output or sink. In convey, all static numbers and strings continuously output, so 5] will continuously output 5 to a dummy output, looping forever.

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1
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Duocentehexaquinquagesimal, 2 bytes

Try it online!

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1
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ABC, 2 bytes

al

l is the actual infinite loop, a is just a placeholder to increment the accumulator at each iteration.

Try it online!

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1
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CSASM v2.4.0.2, 26 bytes

func main:
.lbl a
br a
ret
end

Endlessly jumps execution back to the label a, which just refers to the br instruction itself.
The ret instruction and end token are needed for the function to compile.

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1
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Splinter, 5 bytes

A{A}A

Try it online!

Interpreter crasher. Defines A as itself, then calls A, with infinite recursion.

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1
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MMIXAL, 10 bytes

Main JMP @

Assembly language. Has jump-to-self. Done. (This compiles down to four bytes of actual code, F0000000.)

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1
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Dis, 0 bytes.

Yes, a blank program:

How it works

  • The Dis language accepts an empty program.
  • Begins with c=0.
  • mem[i]=0 for each i, so every cells stand for NOP.
  • Unlike Malbolge, the Dis language does never modify the program by default.
  • This goes forever.

Since TIO doesn't have Dis by native, I made the interpreter. Try it online!

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1
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yuno, 2 bytes

1¿

This will freeze your browser window so I took the TIO link out.

 ¿    While
1     1
      (1)

The 1-byter ɫ (call this link) dies to recursion which unfortunately outputs to STDERR. I might add a flag to disable that in the future.

This will keep applying 1 to 1 until 1 is not true and then print the result after infinite time. Hopefully that's acceptable. If not, you can add the d flag to disable implicit output, or do 1¿ɳ“” - evaluate 1¿, then take ignore that and take the right argument to ɳ, which is “”, an empty string.

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1
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Mascarpone, 8 bytes

[:!]v*:!
[  ]v*   // define and push a new operation
 :!   :! // duplicate and execute the operation on the stack
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1
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Maze, 7 Bytes

^^
%R%L

Maze is a 2D language where cars navigate through a maze and execute instructions in cells they visit. This code starts a car ^^, which by default moves downwards. The next instruction redirects the car to the right %R. That lands on a cell that redirects the car to the left %L, creating an infinite loop.

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1
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Binary Pi Calculus, 6 bytes

0x73 0xB8 0x1B 0xDB 0xB0 0x00

Unfortunately, I could find no existing BPC interpreter. If one does exist, please let me know.

Explanation

0x73     0xB8     0x1B     0xDB     0xB0     0x00
01110011 10111000 00011011 11011011 10110000 00000000
01110011101110000001101111011011101100000 0000000
011                                       Run the following two processes in parallel
   100    1110                            Write the free variable 0...
      1110                                ...to the free channel 0
              000                         End first process
                 001                      Run the following on an infinite number of threads:
                    1011110               Read from the free channel 0 to the bound variable 0
                           110    1100    Write the bound variable 0...
                              1110        ...to the free channel 0
                                      000 End second process

The corresponding pi-calculus program is

a<a>|!a(a).a<a>

Reads and writes are blocking, so each thread will be executed sequentially. As there are an infinite number of them, the program will therefore never halt.

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1
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ErrLess, 0 bytes

An empty program in ErrLess is an infinite loop! This is because if the IP reaches the end of the program it loops back to the start, so programs have to be explicitly halted using ..

ErrLess is a stack-based language I made for fun over the last few months. You can read the docs here, and I also started a tutorial.

Try it online!

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1
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Exceptionally, 1 byte

=

Attempt This Online!

All programs in Exceptionally run as infinite loops. (Unfortunately, the empty program exits immediately instead of looping.) All we need is a no-op to execute each time.

Here, we're using =, which (given no arguments) asserts that the register is equal to itself. In fact, any of the characters +-*^=IF{}! should work. Several others (&UDSW) would also work, except that they will eventually run out of memory and print something to stderr.

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1
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makina, 2 bytes

><

That's it, that's the program.

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1
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Fig, \$\log_{256}(96)\approx\$ 0.823 bytes

(

Beats all y'alls' one-byters mwahahaha

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1
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Rattle, 3 bytes

[]0

Try it Online!

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1
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TacO, 2 bytes

@"

TacO sees the source code as an infinite plane. When it encounters a ", it searches forever for the matching quote, idling forever. @ is required as a program root.

Try it online!

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1
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Arturo, 9 bytes

whileø[]

Try it

This also works and is the same length:

f:$[][f]f
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1
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Thunno, 1 byte

Actually \$ 1 \log_{256}(96) \approx \$ 0.82 bytes but the leaderboard doesn't take decimals.

[

Attempt This Online!

Just a simple infinite loop.

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1
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Swift, 12 bytes

while true{}

Try it online!

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1
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(,) 8 Chars or 1.585 Bytes

(,,,,())
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1
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Desmoslang Assembly, 0 Bytes

It's an infinite loop no matter what (at least until I implement a halt instruction (which I might not even do))

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0
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HPPPL, 16 bytes

while 0=0 do end
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0
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C++ Template Meta-Programming, 48 bytes

template<int n>class S{enum{v=S<n-1>::v}}S<1>::v

There is an infinite loop in the template, so this doesn't compile. Other stuff that doesn't compile, but doesn't need to for the template to recurse (S being private, S<1>::v being free floating)

Can be run at http://coliru.stacked-crooked.com/ for error g++: internal compiler error: Segmentation fault

g++ -std=c++14 -O2 -Wall -pedantic -pthread -ftemplate-depth=162345 main.cpp && ./a.out
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3
  • \$\begingroup\$ Not sure if that is compliant with the rules as it doesn't run out of memory when it stops. If you want to do that, there is the -ftemplate-depth compiler option in gcc that allows you to set the recursion depth to something very high so you'll at least get a segfault out of it. \$\endgroup\$
    – iFreilicht
    Oct 16, 2015 at 14:23
  • \$\begingroup\$ @iFreilicht Yeah, I suppose you're right. I couldn't (be bothered to...) find a online complier that supported that. \$\endgroup\$
    – Nathan
    Oct 16, 2015 at 14:33
  • 1
    \$\begingroup\$ coliru.stacked-crooked.com \$\endgroup\$
    – iFreilicht
    Oct 16, 2015 at 14:35
0
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Univar, 8 bytes

($,,)$,,

Self-calling function.

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2
  • \$\begingroup\$ Seems too recursive to me... is there any tail-call optimization? \$\endgroup\$ Sep 25, 2016 at 13:57
  • \$\begingroup\$ @EriktheGolfer Not at my computer right now, but I believe that the interpreter internally uses a variable-length list for the call stack. \$\endgroup\$ Sep 25, 2016 at 14:03
0
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x86 Machine code, 2 bytes

Machine Code

     00 01
0000 73 FE

Asm source

[SECTION .text]
[bits 16]
[org 0x100]

EntryPoint:
    jnc     EntryPoint

Both the jmp and the call instructions use a 1 byte op-code followed by a 2 byte absolute offset. Conditional jumps on the other hand, are limited to a destination within [+127, -128] bytes of the branch. This is because they are encoded with a 1 byte op-code and a 1 byte relative offset. Dos already cleared the flags before invoking our program, so we know the carry flag will be clear and facilitate an endless (non-crashing) loop. We wont run out of stack-space, which would result in a stack-overflow error, or overwriting and ultimately crashing, our program. Approaches using call will suffer this after about 32,638 iterations. (stack is initally 0xFFFE, (3 byte) program begins at 0x100, each iteration decrements the SP by 2.)

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0
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ROOP, 1 byte

1

At the beginning, the number 1 is converted to an object that contains the number. As there is no operator and the object has nowhere to move, it always happens the same: nothing. The program does not end because there is an existing object.

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1
  • 2
    \$\begingroup\$ An explanation would make this a more complete answer. \$\endgroup\$
    – Alex A.
    Dec 24, 2015 at 3:29
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