152
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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

395 Answers 395

1
4 5
6
7 8
14
2
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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.

| improve this answer | |
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2
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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

| improve this answer | |
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2
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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.

| improve this answer | |
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2
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Python 2, 50 39 bytes

Don't know why nobody mentioned I could just write input()

i=input()
while(int(i)):print 1
print 0

Very simple, if the input is 1, will continuously print 1 and cannot reach the print 0. If it is 0, the while will never fire.

Try it online!

| improve this answer | |
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  • \$\begingroup\$ You can save a few bytes by using stdin instead of command line args. \$\endgroup\$ – pppery Aug 6 '17 at 23:04
  • \$\begingroup\$ 29 bytes, you don't need to cast the input to an int because the input function in Python 2 does eval on the input before returning. You also don't need parentheses around the input call. \$\endgroup\$ – LyricLy Sep 29 '17 at 22:02
  • \$\begingroup\$ Oh, actually... This code doesn't even work. Theoretically, it would work if infinite 1's were given as input, but that doesn't meet the spec. Since while is calling the input function every time, it will just print one 1 and then reach the end of the input and error. Your earlier solution with command line args did work, or you can assign the input to a variable before looping. \$\endgroup\$ – LyricLy Sep 29 '17 at 22:13
  • \$\begingroup\$ @LyricLy yeah, I tried that once. Thanks! \$\endgroup\$ – Stan Strum Sep 30 '17 at 19:14
2
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JavaScript, 26 bytes - Variation of Solomon Ucko's answer

I find it weird how I can't comment, but I can edit his answer...

for(;alert(i=prompt())|i;)
| improve this answer | |
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2
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Forked, 22 21 bytes

-1 thanks to BMO

v >%&
$ |
>-:
  |
 %<

The IP path looks like this if the inputted integer is truthy:

v
|
>-v
  |
 %<

Upon hitting % (print as integer), it goes off the edge of the playing field and wraps around, running >% infinitely.

It takes this path if the inputted integer is falsy:

v >%&
| |
>-^

It prints % 0 and then exits &.

| improve this answer | |
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2
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ORK, 395 bytes

There is such a thing as a m.
A m can p a word.

When a m is to p a word:
I have a scribe called W.
W is to write the word.
I have a linguist called C.
C's first operand is the word.
C's second operand is "1".
C is to compare.
If C says it's equal then I am to loop.

When this program starts:
I have a inputter called R.
I have a word called N.
R is to read N.
I have a m called M.
M is to p N.

Try it online!


Here's the ungolfed version, which is not a whole lot different:

There is such a thing as a truth machine.
A truth machine can process a word.

When a truth machine is to process a word:
There is a scribe called Keymaker.
Keymaker is to write the word.
There is a linguist called Chomsky.
Chomsky's first operand is the word.
Chomsky's second operand is "1".
Chomsky is to compare.
If Chomsky says it's equal then I am to loop.

When this program starts:
I have an inputter called Dave.
I have a word called Input.
Dave is to read Input.
I have a truth machine called Hal.
Hal is to process Input.

The first paragraph defines a truth machine class with one member function, process, which takes a string (i.e. word).

The second paragraph defines process: write the string out, and then compare it against "1". If it's equal, loop.

The third paragraph defines our main function: read a string in, instantiate a truth machine, and have the machine process the string.

Simple, really.

| improve this answer | |
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2
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Stax, 5 4 3 bytes

Crossed out 4 is still 4 ;(

wQc

Try it at staxlang.xyz!

 w      Until popping results in a falsy value:
  Q       Peek and print with a newline.
   c      Copy the value atop the stack.

I'm loving this new language.

Thanks to @Weijun Zhou for reminding me of implicit input.

| improve this answer | |
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  • \$\begingroup\$ I am glad to find another user of Stax writing impressive answers. Maybe you should check out the chat as well although there are not much to see at the moment. \$\endgroup\$ – Weijun Zhou Mar 3 '18 at 20:57
  • \$\begingroup\$ You don't need the comma at the beginning. There is implicit input. \$\endgroup\$ – Weijun Zhou Mar 3 '18 at 21:03
  • \$\begingroup\$ Also I suggest you use this link: staxlang.xyz/#c=wQc&i=0%0A1&a=1&m=2. It automatically runs your program. \$\endgroup\$ – Weijun Zhou Mar 3 '18 at 21:14
  • \$\begingroup\$ @WeijunZhou Whoops, thanks! \$\endgroup\$ – Khuldraeseth na'Barya Mar 3 '18 at 21:14
2
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Reflections, 26 bytes

  _:#_: _<
/#_v     /
\: /

Test it!

Explanation:

The _ at position (2|0) reads a line from input, : doubles the first character. Then, # redefines zero and _ at (1|0) prints that character. : doubles the first character again, _ at (4|0) converts a digit to a number. < tests this number, if it's 0, the IP is directed upwards (out of the program). Else it's directed downwards, then left by the /, and down again by the v, left by the /. The : doubles the character again. Then, the \ reflects the IP upwards and the / right again, where # redefines zero and the _ at (1|0) prints the character. The v then directs it down into the loop again.

| improve this answer | |
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2
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Python 2, 59 51 49 33 bytes

i=input()
while i:print 1
print 0

Explanation:

input()             # take input from STDIN
while i:print 1     # print 1 if the argument is anything other than 0... input() evaluates the string and returns a value
print 0             # print 0 and exit if input() returns 0

EDIT: Saved 8 bytes thanks to @EsolangingFruit

EDIT 2: Coupla more bytes thanks to @HyperNeutrino

EDIT 3: Saved 16 bytes thanks to @HyperNeurtrino

| improve this answer | |
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  • 1
    \$\begingroup\$ You don't need to have str(1), you can just do print"1", etc. \$\endgroup\$ – Esolanging Fruit Mar 6 '18 at 2:57
  • \$\begingroup\$ remove the space between print and ' (this is Esolanging Fruit's suggestion but you added an extra space so credit them instead for 8 bytes) \$\endgroup\$ – HyperNeutrino Mar 6 '18 at 3:28
  • \$\begingroup\$ also, don't take arguments, use input. k=input()>"0" and then while k:print 1 and then print 0 (note that you don't even need to do print"1", just do print 1 \$\endgroup\$ – HyperNeutrino Mar 6 '18 at 3:29
  • \$\begingroup\$ @HyperNeutrino Thanks. Fixed the first thing... still looking at input() it's not doing what I expect it to. \$\endgroup\$ – Allen Fisher Mar 6 '18 at 4:42
  • \$\begingroup\$ @HyperNeutrino This is Python 2, so input() returns an evaluated value instead of a string. This means that input() already returns the number. \$\endgroup\$ – Esolanging Fruit Mar 6 '18 at 5:06
2
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MATL, 4 bytes

`GtD

Relevant MATL features *

To explain the code, the following MATL features need to be presented first.

  • ` ...] is a "do ...while" loop. The top of the stack is consumed at the end of each iteration, and used to decide whether to go on with a new iteration or not. The last ] can be omitted if it's at the end of the program (loops are implicitly closed).
  • G works as follows:
    • When there has been no user-input it does nothing;
    • When there has been one user-input it pushes it onto the stack
    • When there has been more than one user-input it takes a numeric argrument and pushes one of those user-inputs onto the stack
  • t by default duplicates the top element of the stack
  • D by default displays the top element of the stack, and consumes it.
  • If a function requires more inputs than currently are in the stack, user-input is implicitly triggered. The entered elements are placed below the current bottom of the stack.

Code explanation

` enters the loop. At the first iteration, G does nothing. t implicitly asks for user input and duplicates it. D displays and consumes the duplicate, leaving the original input on the stack. If this input is 0 the loop is exited and the program finishes (a single 0 has been displayed). If the input is 1, control goes to the beginning of the loop again. Now G pastes the input, t duplicates it, D displays and consumes that duplicate, and again there's a 1 to be used as loop condition, so the loop begins again, indefinitely (an infinite number of 1 is displayed).


* at the time of writing. The behaviour of G has changed since then: G now triggers implicit input when there has been no user-input yet. However, this doesn't affect the code, which works the same. The only difference is that in the first iteration the implicit input is now triggered by G, not by t.

| improve this answer | |
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2
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Pepe, 15 bytes

REeErEErReEEree

Try it online!

Explanation:

REeErEErReEEree  # Full program
REeE             # Take input as number
    rEE          # Create label 0
       rReEE     # Output number -r (preserve)
            ree  # Goto label 0 if [input] != 0
| improve this answer | |
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2
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Self-modifying Brainfuck, 30 19 bytes

Similar method to my BF answer. Input and print, subtract the 0 (the source's last byte) from the input, so cell is 0 or 1. Loop printing if input cell is 1. Tested on my Python interpreter.

,.<[->-<]>[<<.>>]10

You can also run it on TIO.

| improve this answer | |
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2
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Flobnar, 10 bytes

<1._
|@&
0

Try it online!

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

AlgiX with -x, 3 bytes

[.]

Explanation

AlgiX takes each character of input (converts it to charcode if it's not a digit) and runs the entire code for each value, initializing the accumulator to that value each run.

In this case, there's only 1 input, a 1 or a 0, so the code runs once.

[.] - Implicit a = input
[   - If a == 0, skip to ]
 .  - Print value of a
  ] - If a != 0, jump back to [
    - Implicit output value of a

Normally, the implicit output after each run would output a as a character, however the -x flag overrides that and outputs its value instead.

| improve this answer | |
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2
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VSL, 36 32 bytes

fn main(){do{print(i)}while i>0}

This is probably familiar to anyone who knows any C-esque language but this takes input using a global variable i.

Try it online!

| improve this answer | |
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2
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Befunge, 4 bytes

&:.%

Try it online!

Please note: this answer is no longer valid with this particular interpreter. At the time of writing, the interpreter used had a quirk where trying to read more integers when there weren’t any would return the last integer. I believe that this was “fixed” in some interpreter update. (Thanks to @osuka_ for realizing that this answer no longer works)

Explanation:

&       Take number input
 :.     Duplicate and print input
   %    Mods the second from-the-top with the top number in the stack
        This means that if the top is 1, it will yield 0 % 1 = 0,
        But if it is a 0, it errors out and stops execution because you are
        trying to divide by 0
        
&       The program wraps back around if it didn't error, and the & takes
        the last number in the input if it has already taken them all, yielding
        the original 1 or 0

Coincidentally, this doesn't also work for /, because it strangely asks you what you want the answer to be.

| improve this answer | |
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  • 1
    \$\begingroup\$ The modulo trick is really clever. I love it. \$\endgroup\$ – IQuick 143 Apr 30 '19 at 14:15
  • 1
    \$\begingroup\$ This doesn't work - it prints -1 instead of 1 after the first time. \$\endgroup\$ – osuka_ Aug 26 '19 at 18:56
  • \$\begingroup\$ @osuka_ I think it’s due to the interpreter being updated, unfortunately. I remember it used to be that reading an end of feed with & would give the last integer again (which would make this print 1). I think they changed it so it gives -1 on an EOF instead. Thanks for letting me know \$\endgroup\$ – MildlyMilquetoast Aug 26 '19 at 19:19
  • 1
    \$\begingroup\$ @MildlyMilquetoast No worries - The reason I stumbled upon this is that I was looking for other befunge answers, to see if anyone had already posted one identical (or very similar) to one I just wrote. I love the mod trick! \$\endgroup\$ – osuka_ Aug 26 '19 at 19:30
  • \$\begingroup\$ This can be fixed by adding two !s after the & \$\endgroup\$ – Jo King Aug 27 '19 at 3:48
2
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Verbosity v2, 290 bytes

IncludeTypePackage<OutputSystem>
IncludeTypePackage<InputSystem>
IncludeStructurePackage<Loops>
input=InputSystem:NewInput<DEFAULT>
outpu=OutputSystem:NewOutput<DEFAULT>
condi=InputSystem:ReadEvaluatedLineOfInput<input>
Loops:ConstructLoop<Do;condi>[OutputSystem:DisplayAsText<outpu;condi>]

Try it online!

First non-trivial Verbosity v2 answer! Although how it works is very trivial, as you can see from the ungolfed version:

IncludeTypePackage<OutputSystem>
IncludeTypePackage<InputSystem>
IncludeStructurePackage<Loops>

input  = InputSystem:NewInput<DEFAULT>
output = OutputSystem:NewOutput<DEFAULT>

condition = InputSystem:ReadEvaluatedLineOfInput<input>

Loops:ConstructLoop<Do; condition> [
	OutputSystem:DisplayAsText<output; condition>
]

Try it online!

Simply enter a do...while loop, which prints the input at least once, then terminates if it's falsey (i.e. 0).

| improve this answer | |
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2
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W s, 1 byte

The s flag forces a read from STDIN. This flag can be appended to the end of the command-line.

w

No, I didn't just make a built-in for the question.

Here is the expanded program (because w requires two inputs)

aaw

This decodes into:

mywhile(a[0],'a[0]',a)

Which means:

while a[0]:
    print(a[0])

The print is implicit in every while iteration. Since a[0] is always 1 once it is defined as 1, this will print 1 forever (the body is the same as the condition).

So you might say that this prints nothing if the input is 0. You are wrong, the w instruction returns the condition 0 after the while statement is done. Therefore it gets returned and gets outputted.

W s, 4 bytes

W was designed to be very powerful so that while loops are unneccecary. This program avoids this at the cost of being significantly longer. (Please don't use the while loop (which is simply for presentation purposes) in your W programs. Thank you.)

ibE&

Explanation

   & % Perform logical AND between the input and the code block.
     % If the input is 1, execute the following code block:
i E  %     Count to infinity,
 b   %     Printing the input in every iteration
     % Otherwise (part of the AND logic):
     %     Return 0, exit the program.
| improve this answer | |
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2
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tq, 3 bytes

?(?

Explanation

?,   # Take an input
  (? # Extend the array while
     # the input is not false
| improve this answer | |
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2
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Shakespeare Programming Language, 123 120 bytes

(Whitespace added for readability)

T.Ajax,.Puck,.Act I:.Scene I:.[Exeunt][Enter Ajax and Puck]
Ajax:Listen tothy.Open heart.
     Am I worse you?If solet usAct I.

Try it online!

I make use of a trick to reuse Scene I. For some reason, it doesn’t error to keep taking input, so I do that. Then I just compare the input to zero, and loop if it’s greater. Saved 3 bytes by comparing to I instead of zero because characters initialize to zero—thanks Robin Ryder!

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  • \$\begingroup\$ You can use Is I worse you? instead of Is zero worse you?, since all characters are implicitly initialized at 0. \$\endgroup\$ – Robin Ryder Jan 21 at 14:59
  • \$\begingroup\$ True true, thank you! \$\endgroup\$ – Hello Goodbye Jan 22 at 15:03
2
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MAWP v1.1, 18 14 bytes

@A{1A:.}1M[!:]

-4 bytes with integer input.

Try it!

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  • \$\begingroup\$ Nice answer! You should specify it is MAWP 0.1 (0.0 is not allowed, by the way, because it is not a programming langauge at all). \$\endgroup\$ – null Aug 5 at 13:51
  • \$\begingroup\$ OH, sorry about that, I just used the answer generator here: 8dion8.github.io \$\endgroup\$ – Razetime Aug 5 at 15:18
2
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MarioLang, 25 bytes

   >:<
   "==
 ;[!:
===#=
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  • 2
    \$\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
  • 1
    \$\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
1
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Ruby, 32 30 bytes

p gets.to_i<1?0:loop{p 1};exit
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1
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APL, 9 bytes

{⎕←⍵:∇⍵}⎕

Explanation:

          ⎕  ⍝ read a number from the keyboard
{⎕←⍵:   }    ⍝ output it, and if it is true:
      ∇⍵     ⍝ call the function again with that input
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1
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Pip, 5 bytes

WaPaa

In pseudocode:

while(a)
    print(a)
autoprint(a)

a is the first command-line argument. (The OP commented on the sandbox post that this was an acceptable way to take input, even though Pip is capable of taking input from stdin.)

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  • \$\begingroup\$ WaPaa. I love these weird types of languages because when you put them into TTS programs they have a conniption. \$\endgroup\$ – Cyoce Dec 16 '15 at 18:29
1
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Stack, 67 bytes

{ '1' print a call } `a set '' input num 1 = a { '0' print } ifelse

Run by placing the stack folder into your python lib folder, and running py -3 -m stack.cli truth.stack

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1
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3var, 6 bytes

">|[w]

Each 0 or 1 is followed by a newline. I've never actually used 3var's loop features before, so this is a first for me.

Explanation

"         Read input into R
 >        Copy R into A
  |[ ]    Do while A > B...
    w     Output R
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1
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Browser LiveScript, 54 bytes

switch prompt!
|'0'=>alert 0
|'1'=>while true
 alert 1

I believe I can use prompt and alert instead of STDIN and STDOUT with browser languages, is that correct?

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  • \$\begingroup\$ Yes, you are correct. \$\endgroup\$ – Conor O'Brien Jan 16 '16 at 18:17
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