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

494 Answers 494

1
11 12
13
14 15
17
1
\$\begingroup\$

Nekomata, 2 bytes

ᶦP

Attempt This Online!

ᶦ    Iterate the following function zero or more times until it fails
 P   Check if the input is positive
\$\endgroup\$
1
\$\begingroup\$

Funky, 28 bytes

s=io.read()while(-print(s))a

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Thunno 2, 2 bytes

Attempt This Online!

    # implicit input
(   # while TOS is not 0:
 ß  #  print without popping
    # implicit output
\$\endgroup\$
1
\$\begingroup\$

Tsept v1.0, 45 bytes

Tsept is an open-source esolang I created; it's not really meant for code golfing, as even basic tasks like setting a register to a number take many instructions, and even though there are seven different registers it's almost impossible to do anything without the stacks.

?PBKpPxIIIPAPAPAPABpBPBSPBxDDPAPADPBKB!BKBPBi

Note that you'll need a Linux machine to run this since there is no web-based interpreter for it (yet).

Analysis

?PBKpPxIIIPAPAPAPABpBPBSPBxDDPAPADPBKB!BKBPBi
?PBKpP                                        Store input in register B
      xIIIPAPAPAPA                            Set accumulator to 48
                  BpBPB                       Swap stack value and accumulator
                       SP                     Subtract ASCII 1 from accumulator and push to stack
                         BxDDPAPADPB          Calculate jump offset
                                    KB!BKBPBi Loop
\$\endgroup\$
1
\$\begingroup\$

(,) 63 Chars or \$63\log_{256}(3)\approx\$ 12.48 Bytes

This was quite fun, kudos to Dadsdy for creating this language!

((),((())))(()(),(()())()(),,,(()),(()()))(,,,(()),(()()),(()))

Try it!

Explanation

Set var_1 to the ascii value of the input (48 or 49)
((),((())))

var_2 = var_2 + 2; repeat while var_1 > var_2
(()(),(()())()(),,,(()),(()()))

print var_1; repeat while var_2 > var_1
(,,,(()),(()()),(()))

In practice this counts up in twos until var_2 becomes at least as big as the input. If the input was 0 (ascii 48) var_2 will be 48 and the last loop will run only once, if the input was 1 (ascii 49) var_2 will be 50 and the last loop will never terminate.

\$\endgroup\$
1
  • \$\begingroup\$ The behaviour of the loop executing only once if the conditions are equal is pretty helpful \$\endgroup\$
    – Jo King
    Jun 23, 2023 at 2:27
1
\$\begingroup\$

(,), 46 bytes

((),(((()))(())),,,(),,((())))(,,(()),,((())))

Try it online!

Outputs separated by newlines.I also have an alternative 46 byter using a similar strategy.

Explanation:

Note that the var shorthand below will return -1 for 0 (representing EOF) and 1 for -1, otherwise the value stored at that index (default 0).

((),(((()))(())),,,(),,((())))   First expression
 ()               Set var1
   ,(((()))(()))  To var(var(var(1))+var(1))
                  So when var(1) maps to:
                    0 -> var(var(0)+0) = var(-1+0) = 1
                    1 -> var(var(1)+1) = var(1+1)  = 0
                  basically alternating between 0 and 1 each loop
                  my other solution flips between 0 and -1 instead
                ,,,(),           Repeat while 1>-1
                      ((()))     A maximum of input char times
(,,(()),,((())))  Second expression
 ,,(())           Output var1
       ,,((()))   Execute once if var(var(1)) == -1
                  And repeat forever if var(var(1)) > -1
\$\endgroup\$
1
\$\begingroup\$

Thue++, 17 bytes

0::=~0
(?=1)::=~1

Input is the state string, output is the alerts.

\$\endgroup\$
1
\$\begingroup\$

MaybeLater, 48 bytes

whenx is1{write(1)when0spass x=1}write(x=read())

One could remove when0spass for 37 bytes, though doing so causes a stack-overflow pretty quickly.

A Pretty Canonical MaybeLater answer, hooks to x being assigned to 1 by writing a 1 and setting x to 1 again, causing an infinite loop. when 0 seconds pass is required to avoid recursion errors by using the event queue instead. Either way, writes the result of the read.

Try it online!

\$\endgroup\$
1
\$\begingroup\$

Shenzhen I/O, 51 bytes, 3¥, 5 Lines

@mov p0 acc
teq acc 0
+mov 0 x1
+mov 0 x0
-mov 1 x1

Takes an input on p0 and outputs to a LCD at x1. If the input is 0, writes to an unconnected XBus wire and will wait indefinitely (this is the closest to terminating that I can think of). If the input is non-zero, loops forever writing 1.

Here is a screenshot of the two cases simultaneously.

enter image description here

\$\endgroup\$
1
\$\begingroup\$

><x>, 105 bytes

lddddddddddddddddddddddddddddddddddddddddddddddddn?v
                                                 >n<
\$\endgroup\$
2
  • \$\begingroup\$ Welcome codegolf @none1 ; This is a nice try. \$\endgroup\$
    – gildux
    Oct 31, 2023 at 17:39
  • \$\begingroup\$ @none1 16 byte version \$\endgroup\$
    – Bee H.
    Nov 29, 2023 at 19:02
1
\$\begingroup\$

><x>, 16 9 bytes

-7 bytes thanks to Bubbler

Abuses the fact that after every instruction, the accumulator is bound to [0, 255]. This allows us to just square the input until it's 0 or 1. Thankfully, ASCII "0", 48, takes only one square to equal 0, and ASCII "1" needs just 4 to equal 1 .

lssss>n?<
\$\endgroup\$
1
  • 1
    \$\begingroup\$ I think lssss>n?< works. \$\endgroup\$
    – Bubbler
    Nov 29, 2023 at 23:58
1
\$\begingroup\$

sed -n, 10 bytes

To avoid the expensive loop, duplicate the line and use D for looping:

h
G
P
/1/D

Try it online!

\$\endgroup\$
1
\$\begingroup\$

ELVM IR, 50 bytes

getc A
mov B,A
sub A,48
.x:
putc B
jeq .x,A,1
exit

Explanation:

getc A     - Input A as ASCII
mov B,A    - Copy A to B
sub A,48   - Subtract A by 48, which converts ASCII to number, while B still stores the ASCII
.x:
putc B     - Output B (The ASCII)
jeq .x,A,1 - If A (The number) is equal to 1, jump to .x
exit       - Exit
\$\endgroup\$
1
\$\begingroup\$

Chef, 179 bytes

a.

Ingredients.
0 a
1 b

Method.
Take a from refrigerator.
V the a.
Put b into 3rd mixing bowl.
Pour contents of the 3rd mixing bowl into the baking dish.
V until ved.

Serves 1.

Try it online!

A test of my new [REDACTED] compiler. As Chef always prints values only at program termination, this technically doesn't fulfill the requirements, but the baking pan (the thing that has the output) does fill up with infinite 1s so it technically counts. Here's the original code as a teaser:

def $in;
read $in;
while $in { print 1; flush; };
\$\endgroup\$
1
\$\begingroup\$

Easyfuck, 5 bytes

"[']'

" - input

[ ] - while loop

' ' - output

\$\endgroup\$
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\$
2
  • \$\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, 2015 at 3:16
  • 2
    \$\begingroup\$ This could be shorter if you just called If io.read()=="1" instead of assigning it to an alias. \$\endgroup\$ Nov 4, 2015 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\$
2
  • 1
    \$\begingroup\$ 2-byte gain: int a=c.Read();do c.Write(a-48);while(a>48); \$\endgroup\$ Dec 6, 2015 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\$ Aug 29, 2016 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\$
1
  • \$\begingroup\$ Can you not remove the spaces around =, after :, and after ,? \$\endgroup\$ Dec 14, 2017 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

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2
  • \$\begingroup\$ ==0 can be replaced with <1 \$\endgroup\$
    – V.G.
    Nov 12, 2015 at 17:18
  • 1
    \$\begingroup\$ val k=readInt;do{print(k)}while(k>0) \$\endgroup\$
    – V.G.
    Nov 12, 2015 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\$
2
  • \$\begingroup\$ Do you need some spaces? ("0" THEN, ? a$) \$\endgroup\$ Jul 22, 2016 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, 2016 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\$
2
  • 2
    \$\begingroup\$ Duplicate of this answer. \$\endgroup\$
    – lirtosiast
    Nov 9, 2015 at 6:02
  • \$\begingroup\$ @lirtosiast Duplicates are allowed. \$\endgroup\$
    – Makonede
    Apr 17, 2021 at 2:45
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\$
3
  • \$\begingroup\$ Actually it's 39 because you have to add either <? or the -r flag \$\endgroup\$ Nov 5, 2015 at 16:11
  • \$\begingroup\$ But you can save one byte more with this: <?for($f=fgetc(STDIN);(print$f)&$f;); - print is like echo but has a return value (always 1) \$\endgroup\$ Nov 5, 2015 at 16:17
  • \$\begingroup\$ @fschmengler Thanks for your feedback. I made it even 2 bytes shorter, by switching (print$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\$ Nov 10, 2015 at 19:15
0
\$\begingroup\$

DStack, 7 bytes

04KKCKT

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

\$\endgroup\$
2
  • \$\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, 2015 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\$ Nov 11, 2015 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\$
9
  • 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, 2015 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, 2015 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, 2015 at 0:42
  • \$\begingroup\$ @Xynariz You should only read the input once \$\endgroup\$ Jan 3, 2016 at 13:35
  • 1
    \$\begingroup\$ @ESP It should still work, I think there's something weird going on. \$\endgroup\$
    – Xynariz
    Jan 12, 2016 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.

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0
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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
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0
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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.

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2
  • \$\begingroup\$ Does it use anything implemented after the question was asked? I think so. Sadly I think this is noncompeting. \$\endgroup\$ Jan 7, 2016 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, 2016 at 11:49
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