152
\$\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

395 Answers 395

1
8 9
10
11 12
14
1
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Pascal (FPC), 54 bytes

var a:word;begin read(a);repeat write(a)until a<1 end.

Try it online!

| improve this answer | |
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1
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Little Man Computer, 19 bytes (4 instructions)

INP
OUT
BRZ 4
BRA 1

Which, when assembled into RAM, is:

901 902 704 601

Annoyingly, BRP means branch if positive or zero, making it useless for distinguishing between the two possible inputs. Therefore we have to use two seperate branching commands: BRZ (branch if zero) to escape the program, followed by a BRA (branch always) to continue the loop if the number is non-zero.

An online interactive interpreter can be found here.

| improve this answer | |
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1
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Backhand, 9 bytes

{I>{@:1O_

Try it online!

A little convoluted, but that's Backhand's way.

Explanation:

{I>{@:1O_
{           # Step left, bouncing off the wall and changing directions
 I          # Get input as a number
  >         # Enter loop by setting the direction to right
     :      # Dupe input
        _   # Step left if 1, otherwise right and change directions
       O    # Both output here, but are going in different directions
    @       # Terminate if 0
  >{  1     # Otherwise push 1 to the stack and restart the loop again
| improve this answer | |
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1
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brainfuck, 28 bytes

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

Try it online!

This should be portable to almost all implementations of brainfuck. If you don't mind negative cells, you can save 2 bytes by removing the initial >>.

Explanation:

 >>                # Go right
 ,.                # Get input and print it once
 [[>]+<[-<]<+>>]   # Convert the number to binary while preserving it
 >[<]<<            # If the last digit of the binary is odd
 [.]               # Print the input forever
| improve this answer | |
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1
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!@#$%^&*()_+, 10 bytes

*0_+!#(!#)

Try it online!

Explanation

*0_+!#(!#)
*             take byte of input
 0            push 48
  _+          subtract
    !#        duplicate and output
      (  )    while the top of the stack is not 0:
       !#     duplicate and output
| improve this answer | |
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1
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Alchemist, 24 bytes

_->In_a+Out_a
a->a+Out_a

Try it online!

I couldn't figure out a shorter way using only one output section. The closest I came was 25 bytes

Explanation:

_->In_a            # a = input
       +Out_a      # Output a
a->a               # While a
    +Out_a         # Output a
| improve this answer | |
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1
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Gol><>, 5 bytes

IZh1n

Thanks to JoKing for helping golf this down further!!!

Try it online!

| improve this answer | |
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1
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Tamsin, 19 bytes

main="0"|{print 1}.

Explanation

The main production first tries to read a 0. If it succeeds, it is returned, and, because this is the main production, printed.

Otherwise, it ignores the input and tries the alternative, which is {print 1}. The { } brackets denote a loop that continues running as long as its content succeeds, and a print statement always succeeds, so this loops forever.

| improve this answer | |
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1
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Python 3, 41 bytes

a=input()
while(a=='1'):print(1)
print(0)

Try it online!

| improve this answer | |
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1
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INTERCAL, 89 bytes

DOWRITEIN.1PLEASE.3<-#1$.1DOCOMEFROM(2)DOREADOUT.1(2)PLEASE(1)NEXT(1)DO(1022)NEXTDOGIVEUP

Try it online!

Looking at some answers to this challenge, I noticed that the current INTERCAL answer was barely golfed at all, with 141 bytes 30 of which are whitespace. Not hard to outdo!

Uses native INTERCAL-72 integer I/O, that is, the input is ZERO or ONE (or some translation thereof) with a trailing newline, and the output is in "butchered Roman numerals".

Explanation:

DO WRITE IN .1

Inputs a number to the 16-bit variable .1.

PLEASE .3 <- #1 $ .1

Sets .3's value to 1 mingled with .1's value. Since .1 holds either 0 or 1, .3 is set to either 2 or 3.

DO COME FROM (2)

If execution would move on normally from statement (2), it goes here instead.

DO READ OUT .1

Prints the value in .1.

(2) PLEASE (1) NEXT

This is statement (2). It places itself on the NEXT stack and moves execution to statement (1). It does not trigger the COME FROM unless RESUMEd to.

(1) DO (1022) NEXT

This is statement (1). It NEXTS to statement (1022) in syslib, which then itself NEXTS to statement (1023) in syslib, which reads PLEASE RESUME .3. This pops a non-zero number of entries equal to .3 off the NEXT stack and resumes execution at the last one popped. Since there are at this point three entries on the NEXT stack, corresponding to (2), (1), and (1022) (RESUME works just fine without labels, it just so happens that all of the NEXTs here have them), and .3 is 2 or 3 depending on whether .1 is 0 or 1, this goes to and moves on from either (1) if .1 is 0, or (2) if .1 is 1. If this RESUMEs to (2), the COME FROM takes effect and begins an infinite loop (and since this outcome leaves the NEXT stack empty, the program does not disappear into the black lagoon after 80 loops).

DO GIVE UP

If (1) was RESUMEd to, execution moves here, and this terminates the program without an error.

(I'll edit in a version with Turing Tape string I/O some time later.)

| improve this answer | |
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1
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NuStack, 82 bytes

putchar(c:char):int;f(i:int):int{if(i>0)putchar('1');else while(1>0)putchar('0');}

Been working on this compiler for a month and a half. Still very much a WIP, but finally at a point where I can do some PPCG with it :D

Unlike all of my other languages, this isn't some toy, esoteric, interpreted language.
It's a serious language that compiles directly to assembly (nasm syntax).

I'll probably opt to get NuStack on TIO once it's notably more feature-complete than it currently is.

Explanation:

Ungolfed version:

putchar(c: char): int;

f(i: int): int {
    if(i > 0)
        putchar('1');
    else 
        while(1 > 0)
            putchar('0');
}

The first line is a function prototype that allows you to call a function defined elsewhere (think C header files)

putchar in particular is currently the only function in libns, the language's standard library. It takes a single char, and outputs it to stdout. libns is currently available for 64bit linux, so it just makes a syscall using sys_write(), and 32bit "raw", ie not running on a kernel or operating system, where putchar will use the int 0x10 interrupt to print the character in TTY mode

NuStack uses post-fix types, similar to TypeScript, and there is no type inference, so types must always be explicitly declared.

NuStack also doesn't have type casts yet, or do..while loops, so this is, as far as I know, the golfiest way to do it.

| improve this answer | |
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1
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GolfScript, 13 bytes

~{.}{.p}while

Explanation

~       # convert STDIN to an integer

{.}     # while loop condition: remains the same--the input
{.p}    # while loop body: print the input
while

This is one of my first golfs!

| improve this answer | |
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1
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Ruby + -np, 16 12 bytes

p 1while~/1/

Try it online!

~/1/ matches the last line of input to the regular expression /1/

Edit: replaced ;p 0 with the -p flag making the answer -np complete, thanks to @ValueInk

| improve this answer | |
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  • \$\begingroup\$ you don't have to count flags any longer; just note in the body which one you've used (as you already have done) \$\endgroup\$ – Giuseppe Mar 5 '18 at 23:14
  • \$\begingroup\$ @Giuseppe thanks, is this the consensus opinion now? \$\endgroup\$ – Asone Tuhid Mar 5 '18 at 23:16
  • \$\begingroup\$ Yeah, see here. \$\endgroup\$ – Giuseppe Mar 5 '18 at 23:18
  • \$\begingroup\$ @Giuseppe so should I call it Ruby + -n? \$\endgroup\$ – Asone Tuhid Mar 5 '18 at 23:28
  • 1
    \$\begingroup\$ Use the -p flag instead to implicit print the zero ;p \$\endgroup\$ – Value Ink Jun 5 '19 at 2:10
1
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Enterprise, 128 bytes

/©©//NDANDA/final disruptive class fdcBit{final immutable void main(){var Money i=read();;;write(i);;;while(i==1){write(i);;;}}}

It's really annoying how the specification isn't very clear, so I had to guess a lot of things.

| improve this answer | |
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1
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Cascade, 10 bytes

?1/
#?
&#|

Try it online!

My first submission with my new language Cascade! Cascade is a tree-like esolang, similar to Pyramid Scheme, but a bit terser and with wrapping boundaries. Ungolfed, this looks more like:

 @ |
 ? /
 #|
 &|
  ^
 # |
 1 |

First, the ? checks if the printed (#) input (&) is positive. If not, it goes left and terminates since it finds no more instructions. If so, it goes right to the ^, which first executes the left (#1, print 1), then goes to the right, which wraps around back to the ^ for an infinite loop. In the golfed code we use another ? which does a similar thing except with the center and the right, since we know the return value of # is always positive.

| improve this answer | |
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1
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MathGolf, 5  4  3 2 bytes

-1 for dropping block marker

q▲

Explanation

Implicit input
Starts a block
q  Output TOS without a newline.
 ▲ Execute the code block while the TOS is true. Executes at least once.
Implicit output

Try it online!

| improve this answer | |
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  • \$\begingroup\$ I think you can drop the Ä since there's nothing before it now. \$\endgroup\$ – Unrelated String Oct 13 '19 at 3:16
1
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Gaia, 3 bytes

:p↺

Try it online!

Explanation

  ↺  While...
:    Duplicating the top of stack leaves a true value on top:
 p   Print the top of the stack

Input is taken implicitly in Gaia. Every iteration leaves the stack empty, but if there is no input left when it attempts to grab some, it simply takes the most recently given input again.

| improve this answer | |
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1
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Keg, 12 8 7 4 bytes

{:|④

Try it online!

Explanation

# Implicit input
{:|  # While the duplicated item is nonzero:
   ④ # Output the item without popping
     # After that, Output the item remaining on the stack and stop the program
| improve this answer | |
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  • \$\begingroup\$ This can now be 8 bytes due to implicit input \$\endgroup\$ – EdgyNerd Aug 10 '19 at 9:30
  • \$\begingroup\$ 4 bytes Try it online! \$\endgroup\$ – Lyxal Dec 15 '19 at 21:11
1
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Python 3, 35 bytes

a=input()
while a:a=int(a);print(a)

This is the best I can do, don't know if it's the optimal solution though.

Explanation

In Python 3, input() stores the input as a string, so even if the user types 0, it will be stored as '0', which is truthy. This allows me to make a kind of do-while loop, which will always run at least once since '0' and '1' are both truthy. However, I then convert a to an integer ('0' becomes 0 and '1' becomes 1). Now, if the user typed 0, it will not run again since the while loop is now while 0 and 0 is falsy. However, if the user typed 1, it will run forever since the while loop is now while 1 and 1 is truthy.

| improve this answer | |
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  • \$\begingroup\$ 34 bytes \$\endgroup\$ – user85052 Dec 26 '19 at 15:17
  • \$\begingroup\$ @a'_' oh nice, didn't know that was a thing in Python 3.8 \$\endgroup\$ – Sagittarius Dec 27 '19 at 1:22
1
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JavaScript, 39 32 Bytes

x=+prompt();do{alert(x)}while(x)

x=+prompt() gets input from user, and transforms it from a string to a number

do{alert(x)}while(x) is a post test loop, so it always alerts X, then executes again if x not = 0

| improve this answer | |
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1
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Comp, 6 or 8 bytes

The first golfing language (sort of :) ) from 2020!

;{^&}&

That could be a mistake. An alternative program is:

;{^£}£

Explanation

;      Take an input from the console as an integer
 {^£}  While that's nonzero, duplicate & print
     £ If that's zero, directly print
| improve this answer | |
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1
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Python3, 40Bytes

a=input()
while int(a):print(a)
print(a)
  • If input is 0 it will be printed once.
  • If input is 1 it will be printed forever.
| improve this answer | |
\$\endgroup\$
1
\$\begingroup\$

RAKIAC machine language, 6 bytes (34 real for asm)

00000100
00000011
00110100
00100001
00000001
00000010

Representing the assembly:

in
s: out
jez v
jmp s
v: hlt
end

Tried out one of my things for fun (definitely not made for golf).

| improve this answer | |
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1
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Turing Machine Code, 19 29 19 bytes

0 0 * * h
0 * 1 r 0

Try it online!

Saved 10 bytes thanks to @Laikoni

| improve this answer | |
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  • \$\begingroup\$ This is doable with just two rules: 0 0 * * h and 0 * 1 r 0. \$\endgroup\$ – Laikoni Feb 26 at 12:54
1
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Arn, 4 bytes

“l○|

Explained

Unpacked: &&[{1

Takes advantage of the fact that Arn's interpreter is written in Javascript. This means the && and || operator can be used as control flow.

  _     % Variable initialized to STDIN. Implied
&&      % Boolean AND
  [     % Sequence
    {   % Block; used to calculate future entries in a sequence
      1 % Literal one
    }   % End block. Implied
  ]     % End sequence. Implied
| improve this answer | |
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  • \$\begingroup\$ One more answer and I'm going to learn Arn. \$\endgroup\$ – null Aug 21 at 13:59
  • \$\begingroup\$ ;) haha, like your esolang 1+ a lot too \$\endgroup\$ – ZippyMagician Aug 21 at 14:00
  • \$\begingroup\$ (Although, common mistake is I created 1+ - Parcly Taxel created it.) \$\endgroup\$ – null Aug 21 at 14:01
  • \$\begingroup\$ Oh wow, you use it so much I assumed it was yours haha. Apologies to the creator then. \$\endgroup\$ – ZippyMagician Aug 21 at 14:03
1
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Poetic, 137 bytes

TRUTHFUL MACHINE
A:I=true,O=false
B:get a number or get a value
C:if falsy,a zero
D:if truthy,a one
D.a:repeat it if value=I
E:suspend it

Try it online!

| improve this answer | |
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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

| improve this answer | |
\$\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
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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.

| improve this answer | |
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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);
    }
}
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  • 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
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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. :-)

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