121
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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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  • 41
    \$\begingroup\$ I've got to start posting programs with a negative byte count to beat all these empty files! \$\endgroup\$ – CJ Dennis Oct 3 '15 at 4:32
  • 3
    \$\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 '15 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\$ – ThisSuitIsBlackNot Oct 6 '15 at 13:05
  • 1
    \$\begingroup\$ @ThisSuitIsBlackNot Yes, file input is allowed. \$\endgroup\$ – Kritixi Lithos Oct 6 '15 at 16:47
  • \$\begingroup\$ Can you print "", an empty string? \$\endgroup\$ – OldBunny2800 Mar 7 '16 at 23:39

477 Answers 477

0
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0815, 7 bytes

}: :#: 

(trailing space)

}:0:#:0

Explained:

}:0:    :Create a label 0
    #:0 :Go to the label 0 if Z is 0h
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0
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Trigger, 5 bytes

 !  !

Generally, any program with an "ABAAB" pattern should work. toggles the ' ' trigger to 1, ! toggles the '!' trigger, and ! jumps to the nearest ! if the ' ' trigger is 1 (which it is).

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0
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Benul, 0 bytes



In Benul, programs are placed in an implicit infinite loop and require special conditions to terminate.

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0
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AlphaBeta, 2 bytes

 N

 is a no-op, and N moves the IP to the value stored in the position register (initially 0) if the third register is 0 (which it is).

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0
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Java, 50 bytes

interface D{static void main(String[]a){for(;;);}}
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0
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Copy, 10 bytes

copy 0 0 1

Copy the current instruction at the next location.

Basically a Core war imp.

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0
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LI, 2 bytes

R0

Usually, LI programs (under the current interpreter) always take in input. No input is commonly represented as falsy input, i.e. 0; however, in order to truly accept "no" input, I need to provide my own input to the Recurse function.

Explanation:

R   Recurse with input:
 0  Literal 0

LI, 1 byte

If we're a bit more lax with the input requirements, simply R will work. It doesn't error if you don't give it an input, but that's simply because, while it interprets the input as invalid (empty), it doesn't try to use its value.

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0
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ABCR, 2 bytes

5x

Explanation: 5 loops while the front member of queue B resolves to a truthy value. The default value for queue B is 1, so the loop continues until it finds its matching x; since there are no operations in the loop, it continues ad infinitum with no output.

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0
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Lolo, 20 bytes

lOLolo loLOlo loloLO

You can also put in l at the end, but it won't ever get called.

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0
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Sinclair ZX-81 Basic, 6 bytes

1 RUN

This would take 6 bytes in RAM (2 bytes for the line number, 2 bytes for the line length, 1 byte for the RUN token and 1 byte for a newline).

The way it works is (essentially) the same as the 10 GOTO 10, the first line in the program causes the program to be executed.

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0
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dotsplit, 4 bytes

jump

jump: pops number from stack, moves back that many commands

When the stack is empty, popping results in zero. Hence, it will jump on to the jump command.

In keeping with design principles, the problems this has will not be addressed, but perhaps a new command will be added, called loop-safe or something

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0
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CMD, 2 bytes

Must be in a file named a.bat or a.cmd

@a

Must be in a bat or cmd file

@%0
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  • 2
    \$\begingroup\$ The first answer counts as 3 bytes: 2 for the code, and 1 for the restriction on the filename \$\endgroup\$ – Kritixi Lithos Nov 12 '16 at 11:37
  • \$\begingroup\$ Why? He's being punished for a restriction imposed on him by the system, which only recognises specific suffices as executable code. This seems extremely unfair, because it's completely out of his control. He would not be punished if he used a simple batchfile-editor which would execute the code for him, and thus this punishment is undeserved. \$\endgroup\$ – z0rberg's Jan 5 '17 at 18:30
0
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Visity, 0 bytes

(no code)

As this answer points out Visity wraps around producing an infinite loop.
Or more conventional (2 bytes):

[]
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0
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Unary, 695 bytes

695 zeros. Program that prints the actual code can be found here.

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0
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7, 3 bits, represented as 1 byte on disk (language postdates challenge)

1

Try it online!

This is fairly simple. The program starts by pushing 7 onto the stack (which is what 1 does); then when the end of the program is reached, the top nonempty stack element is executed, without removing it from the stack (although any empty elements above it are removed). 7 just pushes an empty stack element to the stack, which immediately gets removed as it's the end of the program, and so nothing has changed and the same stack element just gets run again, repeatedly.

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0
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Pari/GP , 9 Bytes

while(1,)             

(dummy text such that SE accepts the answer)

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0
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PKod - 1 byte

Version 1.1 was released very recently as I still update this, which implements this new function/command

<
Explanation: Jumps back to first byte in code. So it jumps back to itself. Normally meant to be used with test cases (e.g.: if prime, jump back to start - iterate all code again, but since thread asks for infinite loop, why not)

Alternative version, 4 bytes:

-+ia
Explanation: -+ , remove then add 1 to value. Thus value keeps jumping from 0 and 1. "i" jumps two blocks back in code until value is next char after "i". Thus until char is 1. Thus until value reaches ascii code of a which is 97. Which never happens
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0
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Chip, 5 bytes

*S

+3 for -w on the command line, which allows the program to run without reading from stdin. Instead, it uses an internal source (generating 0x00 repeatedly by default).


* is a constant high signal; this turns on adjacent elements.
S when powered, suppresses output for the current cycle.

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  • \$\begingroup\$ I believe the first flag (-w) is considered 3 bytes (but the following flags will be considered 1 byte each) \$\endgroup\$ – Kritixi Lithos Jan 30 '17 at 17:12
  • \$\begingroup\$ You are right! I wasn't certain to start with, and your comment forced me to do research on the accepted norms. \$\endgroup\$ – Phlarx Jan 30 '17 at 17:22
  • \$\begingroup\$ Do you need the -w flag? The consensus is that you can assume STDIN is empty. \$\endgroup\$ – Esolanging Fruit Jul 10 '17 at 0:08
  • \$\begingroup\$ @Challenger5 The problem here is that, by default, Chip terminates when it has processed all of stdin. Empty stdin means that Chip will terminate immediately. \$\endgroup\$ – Phlarx Jul 10 '17 at 15:14
0
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SmileBASIC, 5 bytes

EXEC.

. is the same as 0.0, so it runs EXEC 0, which executes the program stored in slot 0, which is the default slot.

Example use:

PRINT TIME$
WAIT 1
EXEC.
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0
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BitCycle, 3 bytes

>1<

Places a 1 bit on the playfield and infinitely moves it back and forth between the > < devices.

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0
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AWK, 1 byte

0

If I've read the rules correctly, this satisfies the constraints. If we run the code without input, as stipulated, it will be in an infinite loop processing its nonexistent input, forever. If you happen to give it some input, it won't print it.

Usage:

awk '0'

Using an END label is equivalent to this, regardless of what is attached to label, as it requires some input in order to be executed.

The smallest code that doesn't look for input would be (in my version of gawk):

BEGIN{for(;;)1}

I tried without the 1 and AWK wasn't happy.

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0
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ezfuck, 2 bytes

+{

Basically equivalent to +[] in BrainFuck.

+ increments the current cell to 1, then { jumps back a single command while the current cell is non-zero.

Non-competing since I only wrote the language 3 days ago.

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  • \$\begingroup\$ The "non-competing" rule has been lifted, which means that this is a competing submission. \$\endgroup\$ – Esolanging Fruit Jul 9 '17 at 23:49
  • \$\begingroup\$ @Challenger5 Oh, thanks. Are you sure? It doesn't say in the question that new languages are allowed. \$\endgroup\$ – Carcigenicate Jul 9 '17 at 23:53
  • \$\begingroup\$ Marking as NC was the default for newer languages, but it was changed after a recent meta discussion \$\endgroup\$ – Esolanging Fruit Jul 10 '17 at 0:00
  • \$\begingroup\$ @Challenger5 Awesome, thank you for pointing me to that page! That's certainly a welcome change \$\endgroup\$ – Carcigenicate Jul 10 '17 at 0:03
0
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MOO, 30 28 26 bytes

while(!suspend(0))endwhile

The suspend(0) is necessary to avoid "task ran out of ticks'

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  • \$\begingroup\$ Do you need the spaces? \$\endgroup\$ – Esolanging Fruit Jul 9 '17 at 23:48
  • \$\begingroup\$ No. I don't. Good point. \$\endgroup\$ – pppery Jul 9 '17 at 23:49
0
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XSLT, 173 bytes

<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"><xsl:template name="a" match="/"><xsl:call-template name="a"/></xsl:template></xsl:stylesheet>

Requires a trivial XML file as input (<root /> will suffice), since XSLT won't actually do anything unless given input. You can test it here. It's possible to remove some of the spaces or the xmlns:xsl attribute and still have it parse, but most XSLT lint programs will reject the shorter versions, so this is the shortest "correct" XSLT script I could come up with that loops forever.

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0
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Unreadable, 25 bytes

'"""""'"""'""""""'"""'"""

Try it online!

Explanation

'"""""                       While X is true, do Y
      '"""                   [X] - Return 1
          '""""""            [Y] - Set variable X to value Y
                 '"""           [X] - Return 1
                     '"""       [Y] - Return 1
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  • 4
    \$\begingroup\$ Explanation? Also, you haven't formatted it correctly, it should be '"""""'"""'""""""'"""'""". \$\endgroup\$ – Esolanging Fruit Jul 9 '17 at 23:48
0
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Newline, 3 bytes

i[]

Same as brainfuck answer. Adds one and loops

[i]

also works

{} 

works only if loops[0]=infinity

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0
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Cubically, 2 1 byte

)

This should not work, but it does due to how the interpreter handles the number of loops. With no jump points, ) jumps to the beginning of the file, which is )... cue infinite loop.

This is the proper version that will work if I fix the interpreter:

()

( can be jumped to if all provided arguments are truthy (none are provided so they all are). ) can jump back to the most recent ( if all provided arguments are truthy. Once again, none provided so it jumps regardless.

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0
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Nhohnhehr, 11 bytes

+-+
|$|
+-+

Try it online!

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0
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jq 1.5, 13 bytes

until(null;1)

this also works but is no shorter:

repeat(empty)
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0
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4, 12 bytes

3.6147281494

Try it online! (Note: The interpreter used by TiO raises an error with this code. This is caused by the interpreter, not the program itself)

Explanation (According to the wiki specifications):

3.         Start program
6 14 72    Set memory cell 14 to 72
8 14       While cell 14 != 0
9          End while
4          End program
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  • \$\begingroup\$ Do you have an interpreter on which this program works? \$\endgroup\$ – FlipTack Nov 16 '17 at 7:36
  • \$\begingroup\$ @FlipTack I think this should work, but I can't test it right now. \$\endgroup\$ – KSmarts Nov 16 '17 at 15:12

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