172
\$\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\$
18
  • 3
    \$\begingroup\$ Assuming any behaviour is fine for all invalid inputs? \$\endgroup\$
    – Cruncher
    Nov 3, 2015 at 17:33
  • 3
    \$\begingroup\$ @Cruncher Yes, the only inputs you should expect to get are 0 and 1. \$\endgroup\$ Nov 3, 2015 at 17:38
  • 5
    \$\begingroup\$ Catalog is borked. \$\endgroup\$ Nov 6, 2015 at 15:18
  • 2
    \$\begingroup\$ Catalog appears to consider Bf and bf to be different languages. \$\endgroup\$ Nov 10, 2015 at 1:13
  • 1
    \$\begingroup\$ Is one allowed to print an infinite number of 1s? That is, not print out 1s forever, but print out an infinite list of 1s all at once? This would be theoretically demonstrated, but the language has no notion of printing more than once. It just generates an output. So is it okay if it runs indefinitely, but would print infinite 1s given infinite time (to finish executing)? \$\endgroup\$
    – AviFS
    Apr 26, 2020 at 1:42

490 Answers 490

1
4 5
6
7 8
17
2
\$\begingroup\$

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.
\$\endgroup\$
2
  • \$\begingroup\$ I believe removing the last paren should still work. \$\endgroup\$ Mar 2, 2016 at 19:42
  • \$\begingroup\$ @CᴏɴᴏʀO'Bʀɪᴇɴ Thanks, it does! \$\endgroup\$ Mar 15, 2016 at 20:08
2
\$\begingroup\$

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!
\$\endgroup\$
3
  • \$\begingroup\$ The last one mostly makes sense, except for He quackpizzled? :) \$\endgroup\$ Mar 16, 2016 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, 2016 at 22:55
  • \$\begingroup\$ @ETHproductions That's a bit better now \$\endgroup\$
    – MickyT
    Mar 16, 2016 at 23:08
2
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

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!

\$\endgroup\$
0
2
\$\begingroup\$

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())
\$\endgroup\$
0
2
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

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);
\$\endgroup\$
2
\$\begingroup\$

Silicon, 3 bytes

I[]

Explanation:

I          Get input
[          While the top of the stack is 1
           Implicit output
]          End while
           Implicit output
\$\endgroup\$
0
2
\$\begingroup\$

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

\$\endgroup\$
2
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ Would putchar(c)&1 work? \$\endgroup\$
    – Dennis
    Aug 29, 2016 at 16:41
  • \$\begingroup\$ Yes, you are right! One byte saved, thanks @Dennis. \$\endgroup\$ Aug 29, 2016 at 16:51
2
\$\begingroup\$

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

\$\endgroup\$
2
\$\begingroup\$

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"));
    }
}
\$\endgroup\$
4
  • \$\begingroup\$ That's 143 characters. \$\endgroup\$
    – clismique
    Sep 1, 2016 at 8:49
  • \$\begingroup\$ Why did you even check? \$\endgroup\$
    – Shaun Wild
    Sep 1, 2016 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, 2016 at 8:54
  • \$\begingroup\$ Ahh ok. It's shorter now anyway :P \$\endgroup\$
    – Shaun Wild
    Sep 1, 2016 at 8:55
2
\$\begingroup\$

Lua, 34 Bytes

repeat print(arg[2])until arg[2]<1

arg[2] contains the first command line argument, (arg[1] contains the filename)

Give an input through the command line, and it shall spam it if it's 1, or once if it's not.

Simple enough.

\$\endgroup\$
2
\$\begingroup\$

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
\$\endgroup\$
2
  • \$\begingroup\$ protip: *subtract \$\endgroup\$ Sep 25, 2016 at 9:22
  • \$\begingroup\$ @DestructibleWatermelon No \$\endgroup\$ Sep 25, 2016 at 9:26
2
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

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.

\$\endgroup\$
2
  • \$\begingroup\$ Could I just try this in half life's console, or do I need the source SDK? \$\endgroup\$
    – Pavel
    Dec 21, 2016 at 1:56
  • \$\begingroup\$ @Pavel I tested it on the TF2 console, so that should be fine. \$\endgroup\$ Dec 21, 2016 at 2:02
2
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

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."

\$\endgroup\$
1
  • \$\begingroup\$ Your second version can also be ƓṄ¹¿, where it reads a line instead of a single character. \$\endgroup\$ Mar 12, 2017 at 15:33
2
\$\begingroup\$

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
\$\endgroup\$
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.
█ ██ █
 ██ █
  ██
\$\endgroup\$
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.

\$\endgroup\$
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.

\$\endgroup\$
1
  • \$\begingroup\$ Alternative 7 byter ?v~ newline !~+ \$\endgroup\$
    – Jo King
    Mar 26, 2019 at 9:38
2
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

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

\$\endgroup\$
2
\$\begingroup\$

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.

\$\endgroup\$
2
\$\begingroup\$

Ly, 7 bytes

n:u[:u]

Explanation:

n        # take input
 :u      # duplicate it and print it
   [     # while the top of the stack is not 0...
    :u   # duplicate it and print it
   ]     # end loop
\$\endgroup\$
1
  • \$\begingroup\$ With the current version on TIO, this works n[:u] since the 0 on the stack prints when it exits. \$\endgroup\$
    – cnamejj
    Apr 16, 2022 at 6:40
2
\$\begingroup\$

6502 machine code (C64), 17 bytes

20 FD AE 20 9E B7 8A 09 30 20 D2 FF C9 30 D0 F9 60

Online demo

This is position independent code, you can enter it at any position in RAM and call it, so it doesn't need a load address. Assuming you put it at C000 (that's the case in the online demo), call it like sys49152,0 or sys49152,1 -- any other input is undefined.

Explanation:

20 FD AE    JSR $AEFD    ; consume comma
20 9E B7    JSR $B79E    ; evaluate number, result in X
8A          TXA          ; transfer X to A
09 30       ORA #$30     ; convert to numeric character
20 D2 FF    JSR $FFD2    ; output character
C9 30       CMP #$30     ; compare with '0' character
D0 F9       BNE *-5      ; not equal? Then jump back to output (-7)
60          RTS          ; return
\$\endgroup\$
2
\$\begingroup\$

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!

\$\endgroup\$
4
  • \$\begingroup\$ You can save a few bytes by using stdin instead of command line args. \$\endgroup\$ Aug 6, 2017 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, 2017 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, 2017 at 22:13
  • \$\begingroup\$ @LyricLy yeah, I tried that once. Thanks! \$\endgroup\$
    – Stan Strum
    Sep 30, 2017 at 19:14
2
\$\begingroup\$

><>, 6 Bytes

::n?!;

Input is assumed to be on the stack. It's duplicated twice, printed, and checked if it's non-zero. If it's non-zero, skip the end program command and loop around.

Note that this does not print any spacing between the 1s.

\$\endgroup\$
2
\$\begingroup\$

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;)
\$\endgroup\$
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