Implement a Truth-Machine

Question

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 code-golf 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>

Answer 1

Brain-Flak, 6 bytes

{(())}

While the top of the stack is non-zero, push 1. Input to the stack is implicit, and output would also be, but if the program doesn't halt, nothing is outputted. However, one is continually pushed onto the stack, and it's the closest Brain-Flak can get to infinite output.

Try it online!

Answer 2

Perl -p, with the following restrictions ppencode, 15 bytes.

• Use only keywords with lowercase alphabets only
• Separate each of them with just a space

print while sin

Try it online!

How it works

• sin(0) is zero
• sin(1) is not zero
• -p is for implicit I/O for line by line

Alternatives

sin can be replaced with one of following keywords:

abs int oct hex

Answer 3

Unlambda, 15 bytes

``c``@?1i `c`|i

Try it online!

Answer 4

ErrLess, 12 10 bytes

Thanks to Aaroneous Miller for saving 2 bytes

q@#@8*1-[.

No newlines/spaces.

Explanation

q     { Input a number a number, `n` }
@#    { Output `n` }
@8*1- { Push `8*n-1` to stack (7 for 1, -1 for 0) }
[     { Jump backwards by that many instructions }
      { Goes to right after input for 7, just before halt for -1 }
.     { Halt }

ErrLess is a stack-based language I made for fun over the last few months. You can read the docs here, and I also started a tutorial.

Try it online!

Answer 5

Brainfuck, 27 25 23 bytes

,.[-[->>]<++<]>>>[+[.]]

I ran it here (Uses negative cells, get around this by adding a << to the beginning)

I used modulus 2 instead of subtracting 48 to differentiate 0 and 1 which cuts down some characters. I also copied and did modulus at the same time.

Old answers:

,.[->+>+<<]>[-[<<]>[.]>-]

,.[->+>+<<]>[-[>>]>[.]<<<-]

Answer 6

StupidStackLanguage, 5 bytes

htxux

Try it online!

Answer 7

Plumber, ²⁸ 20 bytes

[]
=]]][=[]
  ][=[][

Try it Online!

I figured Plumber could use some love, so I thought I'd give it a try. It's a pretty fun language to work with! There's no online interpreter for it, so the TIO link is to Javascript, with the interpreter in the header, compliments of @RedwolfPrograms.

This simply takes in input, prints it, and if it is non-zero, it keeps printing it forever.

Answer 8

K (ngn/k), 13 bytes

{x}{ \x}/.1:0

(Don't) Try it online!

The web interpreter does support stdin (via a popup prompt), and entering a 0 will print a single 0. But entering a 1 will hang the entire tab.

We need an unbounded loop, which means either m m/ or a recursive function. The latter being {x o/ \x}@.1:0 is 1 byte longer than the above.

{x}{ \x}/.1:0  full program
          1:0  read stdin as a single list of bytes (= single string)
         .     eval it
{ }{   }/      while loop:
 x               while the value is truthy,
     \x          debug-print the value (print with newline) and return it unchanged

If the input is 0, the loop runs zero times and the resulting value 0 is implicitly printed. If the input is 1, the loop runs forever, printing 1\n repeatedly.

Answer 9

Piet + ascii-piet, 12 bytes (2×6=12 codels)

tavnsJ dl jj

Try Piet online!

As a grid:

tavnsj
 dl jj

As an image:

There was an existing answer with 12 codels, but I got this independently. Any smaller image size is likely not possible, though ascii-piet encoding allows to omit trailing black cells on each line (which makes the linked answer 11 bytes) and therefore may allow for a bit more optimization.

How it works

If the input is 1:

      InN        [1]
Loop: DP+        []   (rotate right once)
      OutN       []   (stack underflow, ignored)
      Push1      [1]
      Dup        [1 1]
      Goto Loop
      DP+        [1]  (rotate right once)
      OutN       []   (print 1)
      Push1      [1]
      Dup        [1 1]
      Goto Loop ...

If the input is 0:

InN    [0]
DP+    []   (go straight through)
Push1  [1]
Not    [0]
OutN   []   (print 0)
HALT

Answer 10

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.)

Answer 11

tinylisp, 36 bytes

This is a language that I'm basing an upcoming challenge on. The spec in the challenge doesn't include the disp function, but the reference implementation does.

(d M(q((x)(i x(i(disp x)0(M x))0))))

Defines a function M that takes an argument x (the closest that the language has to input). If x is falsey, we return 0, which is printed. If x is truthy, we want to display x and then recurse. There isn't any equivalent to Common Lisp's progn in the language, so the best way to do this is to use the disp call as the condition of an if. The result is falsey, thus putting the recursive call (M x) in the else branch.

Answer 12

Sisi, 22 bytes

Sisi doesn't have any way to take user input, so the number is expected to be stored in the variable x (presumably on line 1).

2 print x
3 jumpif x 2

Pretty straightforward: print the number, and keep doing so as long as it's 1.

Answer 13

Stuck, 9 Bytes

Stuck has a while loop function (which I never added to the docs on Esolangs, but it has existed for a little over 2 months) which makes this possible. It wasn't before this as far as I know :P.

ip"1="'ph

Answer 14

Beam, 25 23 13 bytes

rSn(`)>@<
H@<

Try it in the online interpreter! (Warning: it may crash your browser with an input other than 0.)

Beam is a 2D language, based on the concept of a beam of light moving through the 2D source code. Beam is oriented around two main memory values: one held by the beam, and one called the "store". Here are the commands used, in order:

• r - Set the beam to the next ASCII code in the input. (48 for 0, 49 for 1)
• S - Set the store to the beam.
• n - If beam != store, point the beam downward. Does nothing the first time.
• ( and ) - If store != 0, point the beam right/left, respectively.
• ` - Decrement the store by 1.

Now, the beam enters (`) from the left, and bounces back and forth until the store reaches 0. If the store's initial value is 48 (0), it will exit to the left, traveling through these chars:

• n - If beam != store, point the beam downward. This time, since the beam is 48 and the store is 0, it does its job.
• < - Unconditionally point the beam to the left.
• @ - Output the beam as an ASCII character. Prints 0.
• H - Halt the program.

However, if the store's initial value is 49 (1), it exits the loop to the right, and runs through this code:

• > - Unconditionally point the beam to the right.
• @ - Output the beam as an ASCII character. Prints 1.
• < - Unconditionally point the beam to the left.
• @ - Output the beam as an ASCII character. Prints 1.
• > ...

...and so on until the end of time (or until your browser crashes).

Thanks to @MickyT for this awesome layout!

P.S. If you want to learn more about Beam, check out and vote on this post!

Answer 15

JavaScript (ES6), 77 73 64 61 bytes

x=prompt()==1;j=a=>{if(x)setTimeout(j,9);console.log(+x)};j()

This program doesn't crash your web browser!

Explanation:

x = prompt() == 1;             // this makes sure input is 1 or not while defining
j = a => {                     // es6 arrow function
  if (x)                        // if x is 1
    setTimeout(j, 9);            // make sure to do this function again in 9ms
  console.log(+x);              // log the number
};
j();                           // call j

Answer 16

Brainfuck, 54 bytes

+++[>++++[>++++<-]<-]>>>,.[>+<<[>>-<<-]>>>+<<-]>[>.<]

Explanation:

+++[>++++[>++++<-]<-]>>> Set a register to 48 (ASCII 0), using multiplication to reduce the byte count (3*4*4), then set the pointer to the next instruction

,. Receive and print a line of input

[>+<<[>>-<<-]>>>+<<-]Set the next register to be a numerical value, and the one after to represent the ascii output value

> Move the data pointer to the numerical register

[>.<] While the numerical register is not 0, print the ascii register

Answer 17

brainfuck, 20 bytes

,[.>+<-[-[>]<++<-]>]

This requires an interpreter that exits with an error upon stepping out of bounds with < (such as this one). Stepping over the left edge in case of an even value was the easiest way I found to do a parity test. I wouldn't be surprised if it could be a couple of bytes shorter.

Answer 18

Batch, 44 bytes

@set/p n=
:a
@echo %n%
@if %n%==1 goto :a

Explanation

• set /p reads from stdin into a variable
• :a is a GOTO marker, because batch does not have while loops

The rest should be obvious: output the variable n and if n is 1, repeat.

Answer 19

VBA, 54 48 Bytes

Sub f(u):Do:Debug.Print u:Loop While u>0:End Sub

Look Guys, VBA can fit on one line(almost) and be hard(ish) to read like all the other Languages.

Debug.Print could be Msgbox but I feel that isn't the Spirit of the challenge and you really don't want Never ending pop-ups

Old Code

Sub f(u):Do:Debug.Print u:If u=0 Then End
Loop:End Sub

Answer 20

FORTH, 44 bytes 39 bytes 31 bytes

Edit:

As suggested by @Nate Eldredge, we can shorten the code, if we allow extra spaces in the output. This program is 31 bytes long:

: P BEGIN DUP . ?DUP 0= UNTIL ;

Sample run:

1 P
0 P

First version:

: P BEGIN DUP 48 + EMIT ?DUP 0= UNTIL ;

Sample run:

1 P
0 P

Explanation:

We expect the value 'b' of 0 or 1 on the stack.

: P       -- beginning a word P                          ( b ) 
BEGIN     -- starting a loop                             ( b )
DUP 48 +  -- creating an ASCII code for the character    ( b  b+48 )
EMIT      -- echoing the character                       ( b )
?DUP      -- dup'ing the value if non-zero               ( 1  1 ) or ( 0 )
0=        -- testing if the value is zero                ( 1 FALSE) or ( TRUE )
UNTIL     -- end of the loop if true                     ( 1 ) or ( - )
;         -- finishing the word

If I take the challenge seriously, FORTH has Standard I/O capabilities, but it is natural in FORTH to take the input from the stack.

If I use the STDIO feature, the code looks like this (44 bytes)

: P KEY BEGIN DUP EMIT DUP 48 = UNTIL DROP ;

Sample run:

P

(Note, that in my environment the standard input was buggy)

Answer 21

MIPS asm, 24 bytes

main:
li $v0, 5  # load read int
syscall  # exec read int, stores value in $v0
move $a0, $v0  # store in $a0
li $v0, 1  # load print int
loop:
syscall  # print $a0
bgtz $a0, loop  # loop while $a0 is greater than zero

Had to learn MIPS recently, might as well do something fun with it.

Answer 22

C, 35 chars

main(c){for(gets(&c);c%puts(&c););}

This is even hackier than cbojar's solution, from which I copied the abuse of the parameter c (int used as char[4]), along with the reliance on little-endian.

puts returns a non-negatve number on success, which (on my Linux/gcc4.8.2) happens to be the number of bytes printed, which happens to be 2. c%2 tests if c is odd, which is true for '1' and false for '0'.

Answer 23

Marbelous, 23

This version doesn't terminate after the zero is printed, but that should be alright:

..}0
../\//
>0
}0
+O
+O

Old version:

}0@0
\\
../\+O
>0\/+O
@0

It doesn't add a newline between each 1, but whatevs.

Answer 24

O, 14 bytes

i{1{1o1}w}{0o}?

O is a work-in-progress language with loads of commands and an interpreter written in "APL-style C", which means incomprehensible code.

i Get input as String

{ Start a CodeBlock (like ruby)

1 Push 1 to the stack

{ Start a CodeBlock

1 Push 1 to the stack

o Pop the stack and print it

1 Push 1 to the stack

} Push the CodeBlock to the stack

w Do the CodeBlock on the top of the stack while the value under it is true. (Pops them both.)

} Push the CodeBlock to the stack

{ Start a CodeBlock

0 Push 0 to the stack

o Pop the stack and print it

} Push the CodeBlock to the stack

? If the 3rd down value in the stack is truthy, do the CodeBlock 2nd down in the stack, otherwise do the CodeBlock on the top. (Pops the first 3 values on the stack.)

Answer 25

Python 2- 64 bytes

x=input()
if x==0:print"0"
elif x==1:
 while True: print"1"

Answer 26

Java, 87 bytes

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

(has output to STDERR, but that should not matter)

Answer 27

Detour, 2 bytes

,~

Try it online!

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

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

Cells wrap around the edges.

Answer 28

HALT, 49 bytes

1 IF '0' 2 ELSE 3
2 TYPE '0';HALT
3 TYPE '1';SET 1

Pretty simple. If input is one go to 3, output 1, set the pointer to 1 so the program never ends. If input is output 0, print, then halt.

Online interpreter (Firefox only)

Answer 29

Shtriped, 33 bytes

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

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

Explanation:

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

Answer 30

BASTARD, 83 bytes

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

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

Explanation:

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

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

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

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