153
\$\begingroup\$

A truth-machine (credits goes to this guy for coming up with it) is a very simple program designed to demonstrate the I/O and control flow of a language. Here's what a truth-machine does:

  • Gets a number (either 0 or 1) from STDIN.

  • If that number is 0, print out 0 and terminate.

  • If that number is 1, print out 1 forever.

Challenge

Write a truth-machine as described above in your language of choice. The truth-machine must be a full program that follows these rules:

  • take input from STDIN or an acceptable alternative
    • If your language cannot take input from STDIN, it may take input from a hardcoded variable or suitable equivalent in the program
  • must output to STDOUT or an acceptable alternative
    • If your language is incapable of outputting the characters 0 or 1, byte or unary I/O is acceptable.
  • when the input is 1, it must continually print 1s and only stop if the program is killed or runs out of memory
  • the output must only be either a 0 followed by either one or no newline or space, or infinite 1s with each 1 followed by either one or no newline or space. No other output can be generated, except constant output of your language's interpreter that cannot be suppressed (such as a greeting, ANSI color codes or indentation). Your usage of newlines or spaces must be consistent: for example, if you choose to output 1 with a newline after it all 1s must have a newline after them.

  • if and only if your language cannot possibly terminate on an input of 0 it is acceptable for the code to enter an infinite loop in which nothing is outputted.

Since this is a catalog, languages created after this challenge are allowed to compete. 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. Other than that, all the standard rules of must be obeyed. Submissions in most languages will be scored in bytes in an appropriate preexisting encoding (usually UTF-8).

Catalog

The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](http://esolangs.org/wiki/Fish), 121 bytes

<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 320px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><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><script>var QUESTION_ID = 62732; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 12012; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, 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: 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.toLowerCase(), 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 > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) 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); } }</script>

\$\endgroup\$
  • \$\begingroup\$ Can we assume that the program halts when the processor finishes executing the written code, for a machine code entry? \$\endgroup\$ – lirtosiast Nov 3 '15 at 16:58
  • 3
    \$\begingroup\$ Assuming any behaviour is fine for all invalid inputs? \$\endgroup\$ – Cruncher Nov 3 '15 at 17:33
  • 3
    \$\begingroup\$ @Cruncher Yes, the only inputs you should expect to get are 0 and 1. \$\endgroup\$ – a spaghetto Nov 3 '15 at 17:38
  • 4
    \$\begingroup\$ Catalog is borked. \$\endgroup\$ – Addison Crump Nov 6 '15 at 15:18
  • 2
    \$\begingroup\$ Catalog appears to consider Bf and bf to be different languages. \$\endgroup\$ – Mooing Duck Nov 10 '15 at 1:13

390 Answers 390

1
3 4
5
6 7
13
2
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Detour, 2 bytes

,~

Try it online!

, will print a value then push it to the next cell.

~ is a filter, so it will push a value IFF it is greater than 0.

Cells wrap around the edges.

| improve this answer | |
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2
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Shtriped, 33 bytes

e :
t :
.
 d :
 i :
 p :
 .
.
p :

This prints the 0 or the infinite stream of 1's without any trailing newlines or spaces.

Explanation:

e :  \ initialize a variable named ":"
t :  \ prompt for integer input, storing the result in :
.    \ define a function named "." that will only return if : is 0 (the next 4 indented lines are part the function)
 d : \ decrement : if : is positive, else return immediately
 i : \ : must have been 1 to reach here and was just decremented, so increment back to 1
 p : \ print :, which we know is 1
 .   \ recursively call ., endlessly looping
.    \ call . initially
p :  \ if . terminated this line will finally be run, printing : which we know is 0
| improve this answer | |
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2
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BASTARD, 83 bytes

{{fi in 0{!b {= {t 0} ‘1’} {(o <> {{fi out ‘1’}{o}}){o}}{fi out ‘0’}}}}

Note: This language is not qualified. Mostly because I'm still designing it. I just wanted to test drive it against some puzzles.

Explanation:

fi in copies the input into the 0 place on the Variable Stack.

!b is a basic if statement to check if its a "1" or not.

If it is, we just use fi out to print our "0".

Otherwise, we define a new function called o that prints a "1", and then calls itself again.

| improve this answer | |
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  • \$\begingroup\$ Welcome to Programming Puzzles and Code Golf Stack Exchange. This is a great first answer, +1. Your language looks interesting; once you've got enough reputation you could get help developing it on Chat. \$\endgroup\$ – wizzwizz4 Feb 19 '16 at 15:05
2
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NTFJ, 27 bytes

:*:##~~~~~#@|########@|($~^

An online interpreter can be found here.

NTFJ is an esoteric programming language, made by user @ConorO'Brien, that is intended to be a Turing tarpit. It is stack-based, and pushes bits to the stack, which can be later coalesced to an 8-bit number.

  • The only pushable values are 0 with ~ and 1 with #.
  • The only manipulations possible to the actual values of the stack are to wrap eight bits into a byte with @, and to NAND with |.
  • The only logic command is IF with (), but combined with JUMP ^, it can be used to create loops.

Thus, it's quite difficult to manipulate values to do your bidding.

How it works

                               Implicit input: byte 48 for 0 or 49 for 1.
:*:                            Duplicate the top item, pop/output, and duplicate again.
   ##~~~~~#@|                  Push 193 and perform NAND.
             ########@|        Push 255 and perform NAND.
                               These two operations change 48 to 0 and 49 to 1.
                       (       If the top item is not 0:
                        $       Pop the top item.
                         ~^     Push 0 and jump to that instruction.
                               This effectively creates a while loop that loops the entire program.
| improve this answer | |
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  • \$\begingroup\$ I believe removing the last paren should still work. \$\endgroup\$ – Conor O'Brien Mar 2 '16 at 19:42
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ Thanks, it does! \$\endgroup\$ – ETHproductions Mar 15 '16 at 20:08
2
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Beatnik, 26 bytes

J ZD ZD JA K ZZZZJ Z JJ MF

Try it online

An explanation

Words    Scores   Action
J        8        Get character value from input
ZD       12       Duplicate TOS
ZD       12       Duplicate TOS
JA       9        Output TOS character
K ZZZZJ  5 48     Push 48 (char 0) onto stack
z        10       Pop 2 from stack and subtract
JJ MF    16 (7)   If not zero skip back 7 words

Of course something like the following, while still not making sense, looks more like what you would expect for a Beatnik program.

Shall falsey determine truths? We **WithoutAWarningAboutMemoryUse** printed infinitely yeas!
| improve this answer | |
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  • \$\begingroup\$ The last one mostly makes sense, except for He quackpizzled? :) \$\endgroup\$ – ETHproductions Mar 16 '16 at 22:50
  • \$\begingroup\$ @ETHproductions Yea, try to find a word that is SORTAWORDLIKEWITHABIGNUMBER and still fits gets difficult :) Will have a play and see what I can come up with. \$\endgroup\$ – MickyT Mar 16 '16 at 22:55
  • \$\begingroup\$ @ETHproductions That's a bit better now \$\endgroup\$ – MickyT Mar 16 '16 at 23:08
2
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Lua, 56 52 Bytes

I know that this answer in 70 Bytes exists, but its author doesn't look like he's updating it when someone point out an improvement.

if io.read'*n'>0then::a::print"1"goto a end print"0"

Old 56 bytes solution

if io.read()=="1"then while""do print"1"end end print"0"

Nothing special here, just using the fact that "" evaluates to true to save a byte on the infinite loop. Parenthesis for functions parameters aren't mandatory when they only take a single string, which is not stored in a variable.

| improve this answer | |
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2
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Julia, 51 45 bytes

readline()>"0"?while 2>1 print(1)end:print(0)

Read a string from STDIN. If it's lexicographically larger than the string "0" then print 1 forever, otherwise print 0 and we're done.

Saved 6 bytes thanks to Sp3000!

| improve this answer | |
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2
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UGL, 6 bytes

il$o:o

Try it online!

How it works:

i       #stack.push(input)
 l  :   #while stack.peek():
  $     #    stack.dup()
   o    #    print(stack.pop())
      o #print(stack.pop())
| improve this answer | |
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2
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Scratch, 12 bytes

Script
I love how Scratch can be so self-explanatory in the right contexts.

| improve this answer | |
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  • \$\begingroup\$ Your list name counts as bytes \$\endgroup\$ – dkudriavtsev Jul 11 '16 at 23:36
2
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PowerShell, 24 bytes

'for(){1}'*"$args"+0|iex

Generate expression as string and pass it to Invoke-Expression (eval).

It casts $args array to string with "", then multiplication sign casts it to int and string for(){1} is repeated int times (1 or 0 - empty string). Then we add 0 to this string, which will be cast to a string also.

Resulting string expression (for(){1}0 or 0) is then piped to Invoke-Expression, resulting either in endless loop outputting 1 or one-time output of 0.

| improve this answer | |
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2
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WistfulC, 141 bytes

This C has seen better days.

if only int n were 0...
wish for "%d",&n upon a star
someday !n... wish "1" upon a star
*sigh*
wish "0" upon a star
if wishes were horses...

Obviously not competitive, but then neither is this language.

Rough translation to regular C:

int n = 0;
scanf("%d", &n);
while (n) {
    puts("1");
}
puts("0");
exit(0);
| improve this answer | |
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2
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Silicon, 3 bytes

I[]

Explanation:

I          Get input
[          While the top of the stack is 1
           Implicit output
]          End while
           Implicit output
| improve this answer | |
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2
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MarioLang, 25 bytes (Non-competing)

   >:<
   "==
 ;[!:
===#=
| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ Why is this non-competing? Just FYI: "Non-competing" generally means "Uses language features that are newer than the challenge" or "Used to be valid but the rules changed". \$\endgroup\$ – James Jun 30 '16 at 15:16
  • \$\begingroup\$ I mean that I am not using this code to win this challenge. but upvotes are free! \$\endgroup\$ – user54200 Jul 1 '16 at 5:32
  • \$\begingroup\$ That's not what non-competing means here. Also, this answer predates yours and is a lot shorter. \$\endgroup\$ – Dennis Jul 22 '16 at 19:57
  • \$\begingroup\$ @Dennis You're correct. \$\endgroup\$ – user54200 Jul 23 '16 at 2:30
  • \$\begingroup\$ @DrGreenEggsandIronMan I was just aware that there was a shorter one and said "non-competing". \$\endgroup\$ – user54200 Jul 23 '16 at 2:31
2
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T-SQL 23 bytes

No STDIN here so a hard-coded variable.

DECLARE @ INT =1; or DECLARE @ INT =0;

and the truth machine is

l:PRINT @ IF(1=@)GOTO l

| improve this answer | |
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2
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C, 41, 40 bytes

main(c){for(c=getchar();putchar(c)&1;);}

Reads a single character from stdin, writes to stdout.

This version is 1 byte longer than feersum's solution, but removes his/her assumptions onstdin.

| improve this answer | |
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  • \$\begingroup\$ Would putchar(c)&1 work? \$\endgroup\$ – Dennis Aug 29 '16 at 16:41
  • \$\begingroup\$ Yes, you are right! One byte saved, thanks @Dennis. \$\endgroup\$ – Stefano Sanfilippo Aug 29 '16 at 16:51
2
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PHP, 34 bytes

<?for($f=fgetc(STDIN);$f;?>1<?)?>0

Wanted to try to get rid of the print but not sure it's worth it since you have to reintroduce the <? tags

| improve this answer | |
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2
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Java, 143 141 125 88 bytes

interface T {static void main(String[]a){System.out.print(a[0]);main(a[0].split("0"));}}

Ungolfed Test Code

interface T {

    static void main(String[] a) {
        System.out.print(a[0]);
        main(a[0].split("0"));
    }
}
| improve this answer | |
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  • \$\begingroup\$ That's 143 characters. \$\endgroup\$ – clismique Sep 1 '16 at 8:49
  • \$\begingroup\$ Why did you even check? \$\endgroup\$ – Shaun Wild Sep 1 '16 at 8:51
  • \$\begingroup\$ I'm using a userscript for PPCG, and it tells me how many bytes there are. (You should totally use it, it's worth it) \$\endgroup\$ – clismique Sep 1 '16 at 8:54
  • \$\begingroup\$ Ahh ok. It's shorter now anyway :P \$\endgroup\$ – Shaun Wild Sep 1 '16 at 8:55
2
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dc, 13 12 bytes

?[pd1=@]ds@x

Run:

dc -f truth_machine.dc <<< "1"

Adding to the diversity of languages used, I present a dc solution that works as follows:

?              # reads the input and pushes it on top of the stack
[pd1=@]ds@x    # stores the macro command [pd1=@] into register '@' and executes it
   p           # prints the value on top of the stack
   d           # makes a duplicate that is pushed on top
   1=@         # pushes 1, pops two numbers and if they are equal, the macro from
               #register '@' is executed (again), thus making an infinite cycle
| improve this answer | |
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2
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Copy, 67 59 bytes

My new esolang :D

getch a
add b a
add b -48
print a
skip b
skip 1
copy -4 0 1

Explanation:

getch a     Take input in variable 'a'
add b a     Set 'b' to 'a'
add b -48   Substract 48 from 'b'
print a     Print 'a'
skip b      Skip the copy if 'a' is not zero
skip 1      ^
copy -4 0 1 Copy the code block from the print to this instruction after this instruction
| improve this answer | |
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  • \$\begingroup\$ protip: *subtract \$\endgroup\$ – Destructible Lemon Sep 25 '16 at 9:22
  • \$\begingroup\$ @DestructibleWatermelon No \$\endgroup\$ – TuxCrafting Sep 25 '16 at 9:26
2
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Crystal, 48 37 36 bytes

y=gets;y=="0\n"&&(p 0;exit);y=="1\n"&&loop{p 1}
y=gets;y=="0\n"&&(p 0;exit);loop{p 1}
y=gets;y=="0\n"&&(p 0;1/0);loop{p 1}

Edit: Shaved off 10 bytes because the post doesn't specify what should happen on invalid input (like 2). If it does and I misunderstood, let me know.

Edit: Shaved off 1 byte by dying with an error instead of normal exit.

Crystal is statically typed, so I couldn't just do gets.chomp (gets can return nil, and nil doesn't have chomp). The alternative was gets.try &.chomp, but that takes much more space than just having the newlines.

In Crystal (and Ruby) you can do something like puts 0 if y=="0\n", however you can also shave off 2 bytes by doing y=="0\n"&&puts 0 as the && operator returns the last object it tests for truthiness.

loop is a method in the standard lib that infinitely runs the block. It's a much shorter way of writing while true;CODE;end.

p prints the result of .inspect on its arguments to the output. Here I abuse it as a shorter puts because for numbers it'll just return the number.

| improve this answer | |
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2
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Valve scripting language, 38 bytes

alias 0 echo 0
alias 1 "echo 1;wait;1"

This defines two commands, 0 and 1. Type 0 into the console for the zero case, and 1 for the 1 case.

| improve this answer | |
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  • \$\begingroup\$ Could I just try this in half life's console, or do I need the source SDK? \$\endgroup\$ – Pavel Dec 21 '16 at 1:56
  • \$\begingroup\$ @Pavel I tested it on the TF2 console, so that should be fine. \$\endgroup\$ – Conor O'Brien Dec 21 '16 at 2:02
2
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Underload, 16 bytes

((1)S:^)~^:^(0)S

Underload has no way to take input from standard input. The most natural way to take input is therefore from the initial stack: () for 1, (!()) for 0 (this is the normal way to represent numbers in Underload).

Here are Try It Online links for 0 and for 1 (be prepared to kill this one quickly; the infinite loop runs very quickly and will spam up your browser window).

This program didn't need much effort to golf; the most idiomatic way to do things is almost the shortest (I just had to be careful not to let the input get buried too far on the stack). It's easiest to read if I translate the code to a hypothetical functional language:

function x(y)
    print(1)
    x(x)
end
x = x^(input)
x(x)
print(0)

The only weird thing happening here is being able to exponentiate functions, but it's a fairly easy-to-understand operation; for example, raising a function f to the power 3 would produce f compose f compose f, i.e. lambda x.f(f(f(x))). Raising a function to the power 1 does nothing (just like if you'd raised an integer to the power 1); and raising a function to the power 0 gives you the identity function (just like raising a number to the power 0 gives you 1). Actually, the integers in Underload are defined in terms of their effect exponentiating functions, rather than the other way round; the operation is fundamental enough to Underload that you use it to construct most flow control.

| improve this answer | |
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2
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Jelly, 3 bytes

Ṅ¹¿

If reading from STDIN is absolutely required:

ƈOḂṄ¹¿`

I'm not sure which since "Jelly's main input method is via command line arguments, although reading input from STDIN is also possible."

| improve this answer | |
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  • \$\begingroup\$ Your second version can also be ƓṄ¹¿, where it reads a line instead of a single character. \$\endgroup\$ – Erik the Outgolfer Mar 12 '17 at 15:33
2
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SmileBASIC, 23 22 bytes

INPUT N@L?N?N/N
GOTO@L

Ends the program with a divide by 0 error.

If this isn't allowed, here's a 23 byte solution:

INPUT N@L?N
IF N GOTO@L
| improve this answer | |
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2
\$\begingroup\$

WireWorld (It doesnt have a scoring method yet :\)

 ████ █ <= this pixel will be the input. if it is a electron head (1), 
█    █     It will loop forever as 1. if its a wire (0), it will do nothing.
█ ██ █
 ██ █
  ██
| improve this answer | |
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2
\$\begingroup\$

Common Lisp, 30

(do((x(read)))((=(print x)0)))

Common Lisp's print function returns the object that was printed. This reads a value from the user, then prints the value until the return value of the call to print returns 0.

| improve this answer | |
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2
\$\begingroup\$

BitCycle, 7 bytes

Golfed 4 bytes off my example program!

?~<
!~+

Provide input as 0 or 1 on the command-line. The -s or -p debug options are recommended, especially when dealing with infinite output.

Explanation

BitCycle is a 2D language that works by moving bits around a playfield. Commands used in this program are:

  • ? puts input bit(s) onto the playfield, moving east
  • ~ duplicates and negates a bit, turning the original right and the negated copy left
  • < sends bits westward
  • + turns 1-bits right and 0-bits left
  • ! outputs bits

The input hits the first ~. A negated copy turns left (north) off the playfield and is discarded. The original bit turns right (south).

At the second ~, the original bit turns west into the ! and is output. A negated copy turns east.

If the original bit was 0, the negated copy is 1; it turns south at the +, goes off the playfield, and is discarded.

If the original bit was 1, the negated copy is 0; it turns north at the + and then west at the <. The 0 hits the first ~ again, where it turns right (north) off the playfield and is discarded. The negated copy (1) turns left (south), leading to an infinite loop.

| improve this answer | |
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  • \$\begingroup\$ Alternative 7 byter ?v~ newline !~+ \$\endgroup\$ – Jo King Mar 26 '19 at 9:38
2
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Aceto, 11 10 bytes

X
p|1
rip^

reads a string and converts it to an integer. | tests for truthiness (1 is truthy, 0 is not) and mirrors horizontally if truthy (in that case moving to the 1). Otherwise, we'll go to 0, which pushes the 0, we print implicit zero and eXit.

If the number was truthy (e.g. 1), we got mirrored to the 1, which pushes a 1, prints, and goes one cell up (^), going into an infinite loop.

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MarioLANG, 11 bytes

;>:[<
=====

Explanation:

;        Get numerical input and save in current cell
 >       Move left (Required to make an infinite loop)
  :      Output the current cell
   [     Skip next command if cell is 0
    <    Move right

Basically, it loops infinitely unless input = 0

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Wise, 3 bytes

[:]

Try it online!

Wise cannot actually output whenever you want, it only outputs the entire stack when the program terminates. So the solution to infinitely output is to simply infinitely fill the stack. When the program eventually halts (never), it will output the stack that, at that theoretical point in time, will have infinite values in it.

Explanation

[:]  Implicit input from command-line arg
[    If last value is != 0..
 :   ..Duplicate last value on stack
  ]  If last value is != 0, jump back to [

Given a non-zero number, will infinitely duplicate the input on the stack, after an infinite amount of time, will terminate and output the entire stack.

Given zero, jumps to the end of the program and immediately terminates, outputting the stack, which contains only the input.

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