154
\$\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

399 Answers 399

1
\$\begingroup\$

Python 3, 35 bytes

n=input()
while 1:print(n);0/int(n)
\$\endgroup\$
  • \$\begingroup\$ Welcome to the site! Nice first answer. \$\endgroup\$ – Redwolf Programs Dec 1 '20 at 13:31
  • \$\begingroup\$ Also, golfing with suggested edits is usually not recommended, I'll leave a comment on xnor's answer for you. \$\endgroup\$ – Redwolf Programs Dec 1 '20 at 13:33
  • \$\begingroup\$ @RedwolfPrograms Thanks for the tip 😅 I'll pay more attention next time \$\endgroup\$ – SeoFernando Dec 1 '20 at 13:40
  • \$\begingroup\$ No problem, it happens a lot. Hopefully you'll get to 50 rep soon! \$\endgroup\$ – Redwolf Programs Dec 1 '20 at 13:44
0
\$\begingroup\$

Lua For windows, 70 bytes

Use lua for windows for this

I=io.read() If I=="1" then while 1 do print"1" end elseif print"0" end

This program works because it takes a 0 or 1 from stdin then if it's a one it makes a while loop that prints one if it's a 0 it prints 0 then the program ends

\$\endgroup\$
  • \$\begingroup\$ Does this rely on functionality specific to Windows? Why couldn't this be run with Lua on other systems? \$\endgroup\$ – Alex A. Nov 4 '15 at 3:16
  • 1
    \$\begingroup\$ This could be shorter if you just called If io.read()=="1" instead of assigning it to an alias. \$\endgroup\$ – James Murphy Nov 4 '15 at 5:41
0
\$\begingroup\$

Ceylon, 91 88 bytes

Improved (explanation at the end):

shared void run()=>("0"<(process.readLine()else"")then{1}.cycled else{0}).each(print);

This was the original (91 bytes):

shared void run(){if("0"<(process.readLine()else"")){while(0<1){print(1);}}else{print(0);}}

Unfortunately the most part goes to reading the input and making sure it is defined.

shared void run() {
    if("0"<(process.readLine() else "")) {
        while(0<1) {
            print(1);
        }
    } else {
        print(0);
    }
}

I merged this null check (in the else "" form) with the decision into one if, so when the input is ended without a line being read, it also prints 0. Also, any string sorting less than "0" lexicographically prints 0, and any string lexicographically > "0" will produce the 1-loop.

The 0<1 is used instead of true because it is shorter.

Here is a functional approach, unfortunately it can't compete (107 bytes after whitespace removal) due to the length of parseInteger:

shared void run() {
    value i = parseInteger(process.readLine() else "") else 0;
    (1:i).cycled.follow(i).each(print);
}

It cycles a range starting at 1 of length i (i.e. either {} or {1}), resulting in {} or {1,1,1,1,...}, prepends i (which results in {0} or {1,1,1,1,1,1,...}), and prints each element.


A different functional approach without variables and parsing is this one (88 bytes, shrinked version at the top):

shared void run() =>
        ("0" < (process.readLine() else "") then
    { 1 }.cycled else { 0 }).each(print);

This does the case distinction in an expression, using the then and else operators to produce either the {1,1,1,...} stream or the {0} singleton, and then calls print(...) on each element of that stream.

\$\endgroup\$
0
\$\begingroup\$

C#, 120 98 bytes

using c=System.Console;class o{static void Main(){if(c.Read()=='1')for(;;c.Write(1));c.Write(0);}}

Ungolfed:

using c=System.Console;
class o
{
    static void Main()
    {
        if(c.Read()=='1')
            for(;;c.Write(1));
        c.Write(0);
    }
}
\$\endgroup\$
  • 1
    \$\begingroup\$ 2-byte gain: int a=c.Read();do c.Write(a-48);while(a>48); \$\endgroup\$ – LegionMammal978 Dec 6 '15 at 16:10
  • 1
    \$\begingroup\$ 95 bytes based on @LegionMammal978 answer using c=System.Console;class P{static void Main(){int a=c.Read()-48;do c.Write(a);while(a>0);}} \$\endgroup\$ – krontogiannis Aug 29 '16 at 10:26
0
\$\begingroup\$

x86 (Linux / NASM syntax), 86 bytes

mov dx,1              ; 3rd argument to read(): length of buffer
mov ecx,esp           ; 2nd argument to read(): the buffer
                      ; 1st argument to read(): the FD is already 0
mov ax,3              ; read system call number
int 128               ; invoke syscall, returns eax=1 == num of read bytes
l:mov ax,4            ; start of loop, eax=syscall number of write()
int 128               ; invoke syscall, returns eax=1 == num of written bytes
cmp byte[esp],49      ; compare buffer against '1'
jz l                  ; loop of it's the same
int 128               ; invoke syscall otherwise (syscall 1 == exit)

Compile with nasm -f elf h.asm && ld -o h h.o -m elf_i386 and ignore all warnings. :-)

\$\endgroup\$
0
\$\begingroup\$

Swift, 187 bytes

import Foundation
let i = Int("\(String(data: NSFileHandle.fileHandleWithStandardInput().availableData, encoding: NSUTF8StringEncoding)!.characters.first!)")!
print(i)
while i>0{print(i)}

Sadly this has to read from stdin if it can :( Would be so much smaller if I could just declare it inside...

\$\endgroup\$
  • \$\begingroup\$ Can you not remove the spaces around =, after :, and after ,? \$\endgroup\$ – Esolanging Fruit Dec 14 '17 at 0:36
0
\$\begingroup\$

Scala, 46 bytes

if(readInt==0)print(0)else while(2>1)print(1)

Kinda uninteresting and obvious, but there we go

\$\endgroup\$
  • \$\begingroup\$ ==0 can be replaced with <1 \$\endgroup\$ – V.G. Nov 12 '15 at 17:18
  • 1
    \$\begingroup\$ val k=readInt;do{print(k)}while(k>0) \$\endgroup\$ – V.G. Nov 12 '15 at 17:24
0
\$\begingroup\$

Atari Basic, 47 bytes

If I remember the syntax correctly, it will be something like this:

1 input a$
2 IF a$="0" THEN end
3 ? a$
4 goto 3

Please correct me if I'm wrong.

\$\endgroup\$
  • \$\begingroup\$ Do you need some spaces? ("0" THEN, ? a$) \$\endgroup\$ – Erik the Outgolfer Jul 22 '16 at 12:35
  • \$\begingroup\$ @EʀɪᴋᴛʜᴇGᴏʟғᴇʀ I'm not sure, i produced this snippet from memory, i did't use atari basic for over 25 years. \$\endgroup\$ – user902383 Jul 22 '16 at 12:41
0
\$\begingroup\$

Python 2, 46 bytes 34 bytes

x=input() 
print x
while x:print x

Basically, it takes in the input, when 1, prints forever (while loop), but if it isn't 1 (when it is zero) it prints and then exits.

This is my first answer! Yay!

\$\endgroup\$
0
\$\begingroup\$

PHP, 33 37 39 29 bytes

Pretty straight forward:

for($f=fgetc(STDIN);$f&print$f;);

Runs from command line:

php truth.php 1

Edits

  • Added 10 bytes as the program must take input from STDIN if possible. Thanks to Martin Büttner for pointing it out. :)
  • Saved 2 bytes by using if and for instead of while. Thanks to BLackhole
  • Saved 2 bytes by placing everything into the for-loop. Thanks to fschmengler.
  • Saved 2 bytes by rewriting the 2nd expression of the for-loop.
\$\endgroup\$
  • \$\begingroup\$ Actually it's 39 because you have to add either <? or the -r flag \$\endgroup\$ – Fabian Schmengler Nov 5 '15 at 16:11
  • \$\begingroup\$ But you can save one byte more with this: <?for($f=fgetc(STDIN);(print$f)&$f;); - print is like echo but has a return value (always 1) \$\endgroup\$ – Fabian Schmengler Nov 5 '15 at 16:17
  • \$\begingroup\$ @fschmengler Thanks for your feedback. I made it even 2 bytes shorter, by switching (print$f)&$f to $f&print$f. There is a general discussion about the byte count here: Is the PHP opening tag mandatory in byte count?. \$\endgroup\$ – insertusernamehere Nov 10 '15 at 19:15
0
\$\begingroup\$

DStack, 7 bytes

04KKCKT

I invented the language after the challenge began, but before I see it.

\$\endgroup\$
  • \$\begingroup\$ Because it's impossible to prove when somebody saw a challenge, claiming that you didn't see a challenge before you created a language normally doesn't make it eligible. However, since this is a catalog challenges, all languages are welcome, even those created after the challenge, so long as they meet our criteria for being a programming language. Just something to keep in mind for the future :) Welcome to PPCG! \$\endgroup\$ – user45941 Nov 11 '15 at 1:06
  • \$\begingroup\$ @Mego I already know that, I wrote that just to mention something (besides I had to write more because the answer was too short). But still thank you! \$\endgroup\$ – DarkPhantom Nov 11 '15 at 1:12
0
\$\begingroup\$

Java, 90 97 bytes

class T{public static void main(String[]a){while(a[0].equals("1"))System.out.println(1);}}

I'm not sure if this counts because it takes 1 or 0 as an argument rather than STDIN. It's been emotional, either way.

Edit: The above does not print zero, the below does:

class T{public static void main(String[]a){do System.out.println(a[0]);while(a[0].equals("1"));}}
\$\endgroup\$
  • 2
    \$\begingroup\$ You can save two characters by using print instead of println. (There was no requirement saying each 1 must be on a new line). Also, the command line can be added easily as while(a[0].equals(new java.util.Scanner(System.in).nextLine()). However, I don't think this meets the requirement, since it doesn't print 0. \$\endgroup\$ – Xynariz Nov 10 '15 at 1:42
  • \$\begingroup\$ Fixed to print zero in 97 bytes. I had to use println because print must just be appending to the string so never actually prints a line for me. \$\endgroup\$ – ESP Nov 11 '15 at 9:45
  • 1
    \$\begingroup\$ "print must just be appending to the string" .. .what? print and println do exactly the same thing, except that println prints the text plus the current OS's line delimiter. See the documentation: docs.oracle.com/javase/8/docs/api/java/io/… and docs.oracle.com/javase/8/docs/api/java/io/… \$\endgroup\$ – Xynariz Nov 13 '15 at 0:42
  • \$\begingroup\$ @Xynariz You should only read the input once \$\endgroup\$ – SuperJedi224 Jan 3 '16 at 13:35
  • 1
    \$\begingroup\$ @ESP It should still work, I think there's something weird going on. \$\endgroup\$ – Xynariz Jan 12 '16 at 15:20
0
\$\begingroup\$

VB.NET, 106 Bytes

Imports c=System.Console
Module m
Sub Main
i=c.readline
1:c.Write(i)
if i=1 then goto 1
end sub
end module

Straightforward, using goto instead of a loop. Compiles with Option Explicit and Option Strict Off, and Option Infer On

\$\endgroup\$
0
\$\begingroup\$

STATA, 31 bytes

di _r(a)
m
do{$a
}while($a)
end

First, get input from the user and store it in global variable a. Then switch to Mata for better commands. Print the value of a and then if it is truthy (i.e. 1), go to the start of the loop. If the end keyword is omitted, unexpected end of file (end statement missing) is output to standard error. I'm not sure if that's okay.

\$\endgroup\$
0
\$\begingroup\$

Staq, 5 chars

'[:]:

Explanation

'         input number, put it on top of the active stack
 [        jump to corresponding ] if the top stack value = 0 or nonexistent
  :       output top stack value as number
   ]      jump back to corresponding [ if top stack value >0
    :     output top stack value as number
\$\endgroup\$
0
\$\begingroup\$

Cardinal, 8 chars

%:> J.jN

Explanation

%                     create pointer
 :                    enter number, put it on top of stack
  >                   redirect pointer to the right
                      NOP
    J                 skip next instruction if top stack value is > 0
     .                output number to STDOUT
      j               skip next instruction if top stack value is = 0
       N              reflect pointer to the left

The original interpreter, written by Madk, is buggy, and some instructions are missing. I fixed the errors and recompiled the interpreter.

You can clone the fixed interpreter and readme from my GitHub repository.

\$\endgroup\$
  • \$\begingroup\$ Does it use anything implemented after the question was asked? I think so. Sadly I think this is noncompeting. \$\endgroup\$ – Yet Another User Jan 7 '16 at 5:15
  • \$\begingroup\$ @YetAnotherUser Cardinal is from 2010. Madk’s original interpreter (in BlitzMax) contains bugs that break the use of Cardinal, and he forgot to implement one or two commands that exist in his language spec. I fixed these errors and recompiled the fixed source in BlitzMax. Now everything works according to his specification. This is a catalog, the date of implementation does not matter. I uploaded my fix on GitHub yesterday, so everyone can enjoy it. The language is 5 years old; I fixed the errors around June 2015, as you can see e.g. here \$\endgroup\$ – M L Jan 7 '16 at 11:49
0
\$\begingroup\$

Mouse16, 10 bytes

?0+[1!4\]0

Asks for input, then coerces it to an int and tests its truthiness. Then, if it's truthy, 1 is printed and the execution jumps to the 4th byte, where 1 is printed again, and so on forever (until the stack overflows), otherwise, 0 is printed.

\$\endgroup\$
0
\$\begingroup\$

Pike, 94 bytes

Looks like C, but isn't C. Is interpreted. Is fast.

int main(){int x=(int)Stdio.stdin->gets();if(x!=0){for(;;){write("1");}}write("0");return 0;}

Ungolfed:

int main() {
  int x = (int)Stdio.stdin -> gets();
  if (x != 0) {
    for(;;) {
      write("1");
    }
  }
  write("0");
  return 0;
}
\$\endgroup\$
  • 2
    \$\begingroup\$ I always want to pronounce Stdio as Studio :) Can you not write an integer to output? \$\endgroup\$ – ETHproductions Jan 22 '16 at 17:26
  • \$\begingroup\$ @ETHproductions before I actually learned C, I always assumed stdio.h was some sorta studio something because of Visual Studio... I dunno. To write a number to stdout as far as I can tell you need to use write("%d", x); because write wants a string or array(char) argument, which is even longer \$\endgroup\$ – cat Jan 22 '16 at 17:29
0
\$\begingroup\$

Boo, 34 bytes

def f(x):
 print x;while x:print x

Boo is a language inspired by Python and implemented in .NET. Its syntax is very similar to Python 2's.

\$\endgroup\$
0
\$\begingroup\$

CJam, 16 bytes

qi1={"1o_~"_~}&0
\$\endgroup\$
0
\$\begingroup\$

AnnieFlow, 6 bytes

010[!"

Here is the decompressed and more readable program:

010101101100000110111111

Breakdown:

010          there are two characters, 0 and 1
1            input is accepted, first symbol popped will be whatever it is
011          there are two stacks, 0 for output, 1 for input
011          1 push occurs when 0 is popped from 1
00           that push is symbol 0 on stack 0 (output 0)
0            pop from the output (halt) when 0 is popped from 1
0011         2 pushes occur when 1 is popped from 1
01           the first push is symbol 1 on stack 0
11           the second push is symbol 1 on stack 1 (replaces the symbol we popped)
1            pop from stack 1, which we just put 1 on, so we loop forever, outputting 1
11           loop forever doing nothing if the input is empty, won't matter

Note: under some interpretations of the language, the trailing 1s can be omitted, and the advanced interpreter does this automatically, also added 0 to the end if there are any trailing 1s. The actual decompression is

01010110110000011
\$\endgroup\$
  • 1
    \$\begingroup\$ With an interpreter to use the whole 8 bits/byte for golfing purposes, this could be an interesting golfing language. \$\endgroup\$ – lirtosiast Feb 3 '16 at 23:07
0
\$\begingroup\$

Pylons, 7 bytes.

izwp,1}

How it works:

i      # Get command line input.
z      # Skip the next instruction if the top of the stack is 0.
wp,1}  # While 1 is truthy, print the stack.
\$\endgroup\$
0
\$\begingroup\$

ForceLang, 52 bytes

set a io.readln()
label l
io.write a
if a="1"
goto l
\$\endgroup\$
0
\$\begingroup\$

Fission, 12 bytes

This is about the shortest I can do at the moment.

*;R?[   X!@J

Try it online

  R           # Create a atom heading right
   ?          # Set atom mass to input char 
    [         # Set direction right
        X     # Clone atom and sending one back
         !    # Print the character
          @   # Swap mass and energy on atom
           J  # Jump (48 for 0, 49 for 1)
*             # Kill program.  Landing spot for 0
 ;            # Kill atom.  Landing spot for 1

The length of this program is quite dependent on the landing spots for the Jump command. There is also a requirement for a bit of space between the [ direction setter and the X linear cloner. Otherwise an extra 0 will get printed. The input must be either 0 or 1 for this to work. The input is not validated.

Originally I was going to put up a 18 byte program which is more forgiving on invalid inputs.

R?@YI@0'L
O0'&[X'1
\$\endgroup\$
0
\$\begingroup\$

Javascript, 33 31 chars

while(alert(x=x||+prompt()),x);
\$\endgroup\$
  • \$\begingroup\$ I'm getting "ReferenceError: x is not defined". \$\endgroup\$ – Adam Dally Apr 16 '16 at 3:36
0
\$\begingroup\$

Common Lisp, 34 32 bytes

(if (= 1(read))(loop (print 1)))
\$\endgroup\$
  • \$\begingroup\$ This appears to be a duplicate of this. \$\endgroup\$ – user45941 Apr 11 '16 at 18:18
  • \$\begingroup\$ Hmmm ... Are you printing the Zero? \$\endgroup\$ – sergiol Jan 31 '17 at 1:06
0
\$\begingroup\$

Turtlèd, 7 bytes

Turtlèd cannot output without halting, so the truth machine writes down infinite ones, like a turing machine's implementation might, onto the grid if input is 1.

!.{1r.}

Explanation:

!                  input into string variable

 .                 write current char of string var (by default the first char)
                   for a single char input, this will be the entire input

  {1  }            while the current char is 1
                   this loop is never run if input is zero, but if it is one, it will run
                   forever, as the last action of the loop writes 1 to the current cell

    r              move right, so that we constantly move right to write ones to infinite
                   cells

     .             write current char of string var (still by default the first char)
                   this is 1 if the loop was run at all. because the loop runs while the
                   current cell is one, this will cause the loop to rerun every time.

       [implicit]  output grid at end of program, program will only end if input isn't 1
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0
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LI, 4 bytes

?iRP

Explanation:

?i       if input is truthy (1):
  RP      Recurse the program with the printed (implied) input
         Else (0 is the only falsy value in LI) return (implied) input, 
              which is implicitly printed.
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0
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ABCR, 7 bytes

iAo7ox

ABCR is a stack-based programming language that allows a few simple operations on any of its three stacks and a single register.

Explanation:

i        Numerical input.  Places result in register.
 A       Queue register value to queue a.
  o      Output peek(a) as a number.
   7     while(register){
    o      output peek(a) again
     x   }
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0
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Dip, 6 bytes

?_1p?0

Body must be at least 30 characters; you entered 19.

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  • \$\begingroup\$ Perhaps you should provide an explanation rather than a space filler... \$\endgroup\$ – FlipTack Dec 22 '16 at 13:23

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