# Shortest infinite loop producing no output

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>

• I've got to start posting programs with a negative byte count to beat all these empty files! – CJ Dennis Oct 3 '15 at 4:32
• 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 – slebetman Oct 5 '15 at 9:36
• Are programs that read and execute their own source code allowed, or does file I/O break the "must take no input" rule? – ThisSuitIsBlackNot Oct 6 '15 at 13:05
• @ThisSuitIsBlackNot Yes, file input is allowed. – user41805 Oct 6 '15 at 16:47
• Can you print "", an empty string? – OldBunny2800 Mar 7 '16 at 23:39

# BRASCA, 3 2 bytes

1J


Try it!

## Explanation

1J - Move the pointer back one character.


# 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.

# Duocentehexaquinquagesimal, 2 bytes

2º


Try it online!

# 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!

C(gcc, platform-independent), 17 bytes Highschooler here, finally something I can do.

main(){for(;;){}}


I read somewhere that this is the least bytes for an infinite loop, although I'm not sure.

• The least bytes is a 0-byte C program compiled with GCC-EFI. – Makonede May 11 at 20:49

# Locksmith, 4 bytes

0156


0 pushes the next digit (1) to the stack, 5 acts as a label, and 6 jumps to the most recent 5 if the top of the stack is non-zero

# Knight, 3 bytes

W1N


Try it online!

# pretty straightforward
WHILE 1
# The official name for this builtin is NULL.
# Because builtins are determined by the
# first character, I can call it NOP.
NOP


## h, 20 bytes

8,2,9,4,9,6,8,2,-1,0


Neither the language (by Nerdaxe) nor this answer (by VilgotanL) are my creations, but I want to publicize them here.

h (link) is a programming language created by Nerdaxe which is quite limited.

Here's an explanation of how the program works through pseudocode similar to subleq assembler:

neg1 2 #acc is now -1
zero 4 #acc is now 1, the variable zero is 1
zero 6 #acc is now 0, the variable zero is 0
neg1 2 #acc is now -1, go back to 2
neg1:-1 zero:0 #variables


# Daoyu, 1 byte

><


Each instruction is one hex, so two instructions fit into one byte (0x69). The < increases the operating level (a number that affects certain commands) and moves the program reader to the beginning of the program, but only if the op level is less than 9. So, it's preceded by the > command, which decrements the op level (to a minimum of zero).

# HPPPL, 16 bytes

while 0=0 do end


# 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

• 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. – iFreilicht Oct 16 '15 at 14:23
• @iFreilicht Yeah, I suppose you're right. I couldn't (be bothered to...) find a online complier that supported that. – Nathan Cooper Oct 16 '15 at 14:33
• coliru.stacked-crooked.com – iFreilicht Oct 16 '15 at 14:35

# Univar, 8 bytes

($,,)$,,


Self-calling function.

• Seems too recursive to me... is there any tail-call optimization? – Erik the Outgolfer Sep 25 '16 at 13:57
• @EriktheGolfer Not at my computer right now, but I believe that the interpreter internally uses a variable-length list for the call stack. – LegionMammal978 Sep 25 '16 at 14:03

# 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.)

## 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.

• An explanation would make this a more complete answer. – Alex A. Dec 24 '15 at 3:29

# Befalse, 7 bytes

A two dimensional language.

Extremely straightforward; the ! unconditionally skips the next instruction and the / and \ reflect intuitively.

!/\
\/


# Etre, 2 bytes

()


The program sets the first cell to 1, then loops while it isn't zero.

# Skull, 8 bytes

for(;;);


Uses a bug in the Skull2C transpiler, where unrecognized expressions are converted directly into C.

# KimL, 14 bytes

A:
ctrl.goto A


For whatever reason, labels must come before the lines that they label.

# Perl 6, 7 bytes

loop {}


but...

> loop{}
===SORRY!=== Error while compiling <unknown file>
Whitespace required after keyword 'loop'
at <unknown file>:1
------> loop⏏{;}


k.

# Perl 5, 9 bytes

while(){}


boooooooooooooooooooooooooooooooooooooooooring.

# Postscript, 6 bytes

{}loop


Infinite loop that does nothing. Ties up 1 core at 100% when run under ghostscript. Interestingly, the ghostscript interpreter keeps accepting input, so it must be running a separate thread. Not sure what would happen if this was sent to a printer, though.

# Oracle PL/SQL 11.2, 30 bytes

BEGIN LOOP NULL;END LOOP;END;/


# Magic, 7 bytes

skip -1


A simple self-GOTO.

# NTFJ, 2 bytes

Try it here!

~^


This is stiflingly simple. ~ pushes a 0 to the stack, and ^ pops N and goes that position in the source code. The simplest unconditional loop in NTFJ is this.

## GNU Make, 16 bytes

f=$(call f)$(f)


The loop is in make itself, not a program it invokes.

# Pylons, 4 bytes.

w,1}


Loops while 1 is true.

# Milky Way 1.6.5, 3 bytes

&{}


See the docs.

# ForceLang, 9 bytes

Works by abusing the new def instruction.

def a a
a


label l
goto l


1v;d
The 1 tells the interpreter to use one-line mode, v starts an infinite loop, and d loops back to the v.