Simple cat program

Question

One of the most common standard tasks (especially when showcasing esoteric programming languages) is to implement a "cat program": read all of STDIN and print it to STDOUT. While this is named after the Unix shell utility cat it is of course much less powerful than the real thing, which is normally used to print (and concatenate) several files read from disc.

Task

You should write a full program which reads the contents of the standard input stream and writes them verbatim to the standard output stream. If and only if your language does not support standard input and/or output streams (as understood in most languages), you may instead take these terms to mean their closest equivalent in your language (e.g. JavaScript's prompt and alert). These are the only admissible forms of I/O, as any other interface would largely change the nature of the task and make answers much less comparable.

The output should contain exactly the input and nothing else. The only exception to this rule is constant output of your language's interpreter that cannot be suppressed, such as a greeting, ANSI color codes or indentation. This also applies to trailing newlines. If the input does not contain a trailing newline, the output shouldn't include one either! (The only exception being if your language absolutely always prints a trailing newline after execution.)

Output to the standard error stream is ignored, so long as the standard output stream contains the expected output. In particular, this means your program can terminate with an error upon hitting the end of the stream (EOF), provided that doesn't pollute the standard output stream. If you do this, I encourage you to add an error-free version to your answer as well (for reference).

As this is intended as a challenge within each language and not between languages, there are a few language specific rules:

• If it is at all possible in your language to distinguish null bytes in the standard input stream from the EOF, your program must support null bytes like any other bytes (that is, they have to be written to the standard output stream as well).
• If it is at all possible in your language to support an arbitrary infinite input stream (i.e. if you can start printing bytes to the output before you hit EOF in the input), your program has to work correctly in this case. As an example yes | tr -d \\n | ./my_cat should print an infinite stream of ys. It is up to you how often you print and flush the standard output stream, but it must be guaranteed to happen after a finite amount of time, regardless of the stream (this means, in particular, that you cannot wait for a specific character like a linefeed before printing).

Please add a note to your answer about the exact behaviour regarding null-bytes, infinite streams, and extraneous output.

Additional rules

• This is not about finding the language with the shortest solution for this (there are some where the empty program does the trick) - this is about finding the shortest solution in every language. Therefore, no answer will be marked as accepted.

• Submissions in most languages will be scored in bytes in an appropriate preexisting encoding, usually (but not necessarily) UTF-8.

  Some languages, like Folders, are a bit tricky to score. If in doubt, please ask on Meta.

• Feel free to use a language (or language version) even if it's newer than this challenge. Languages specifically written to submit a 0-byte answer to this challenge are fair game but not particularly interesting.

  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.

  Also note that languages do have to fulfill our usual criteria for programming languages.

• If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck derivatives like Headsecks or Unary), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.

• Unless they have been overruled earlier, all standard code-golf rules apply, including the [http://meta.codegolf.stackexchange.com/q/1061](loopholes that are forbidden by default).

As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf; these are still useful to this question as it tries to compile a catalogue as complete as possible. However, do primarily upvote answers in languages where the author actually had to put effort into golfing the code.

Catalogue

The Stack Snippet at the bottom of this post generates the catalogue 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: 290px; 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 = 62230; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 8478; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "//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 "//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(42), 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

TECO, 4 bytes

<^T^T>

^T is CTRL+T. It should be self-explanatory how this works.

For those of you who can't read this infamously write-only language, it simply loops over the input (<>), reading (^T) and immediately printing (^T) bytes.

Answer 2

PHP, 20 bytes

<? fpassthru(STDIN);

Kind of surprised there isn't one for PHP yet...

Answer 3

Emmental, 20 bytes

;#44#46#35#57#63#9!<tab>

The program ends with a tab character. This beats the example on the wiki page by redefining and invoking the tab character (ASCII 9) instead of the asterisk (ASCII 42). The spec doesn't really mention what , does at EOF, but the reference implementation uses getChar, which raises an IOError when the file ends. (The , is "quoted" as #44 in the above program.)

Answer 4

Java, 120 bytes

The other Java solution is shorter than mine but, it uses Apache libraries to get the job done. I think it's worthwhile to put forward a java solution that relies solely on the java standard libraries.

This code works for infinite input and for input that contains null bytes.

class C{public static void main(String[]a)throws Exception{byte[]b={0};while(System.in.read(b)>=0)System.out.write(b);}}

Ungolfed

import java.io.IOException;

public class Cat {
  public static void main(String[] args) throws IOException {
    byte[] input=new byte[1];
    while(System.in.read(input)>=0){
      System.out.write(input);
    }
  }
}

Answer 5

Groovy, 39 bytes

System.in.eachByte{System.out.write it}

Since I wrote a proof-of-concept to confirm that Groovy can handle infinite stream, might as well post it as an answer.

Usage:

groovy <script_name>

Answer 6

Emotinomicon, 29 bytes

⏪⏫🆙⏬😅➕⁉️⏩

Most of the program is just checking for EOF.

Answer 7

LabVIEW, 3 LabVIEW Primitives

Pretty obvious what it does I would say.

Answer 8

PlatyPar, 0 bytes

Try it online

At the beginning of the program, input is implicitly pushed to the stack. At the end of the program, the stack is implicitly printed.

Answer 9

Emotinomicon, 5 chars / 15 bytes

⏫⏪⏬⏫⏩

Press "cancel" to exit the loop. Try it here!

⏫   ⏪   ⏬   ⏫   ⏩   explanation
⏫                   take one character as input, push it to the stack
    ⏪               open loop
        ⏬           pops and outputs top of stack as character
            ⏫       take one character as input, push it to the stack
                ⏩   close loop

Answer 10

NTFJ, 4 bytes

(*~^

An online interpreter can be found here.

NTFJ is an esoteric programming language, made by user @ConorO'Brien, 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 bytes of the input are pushed to the stack before the program is run, with the first byte on top. This particular program is fairly simple:

(      If top of stack is not 0:
 *      Pop byte, output as a character.
  ~     Push 0.
   ^    Pop bit/byte and jump to the instruction at the corresponding index.
        This moves us back to the beginning of the program, which is thus looped until the stack is empty.

Answer 11

Gogh, 0 bytes

This is functional in a Gogh program with input.

---

Usage

$ ./gogh <standard-flags> <code-or-path> <input>

In this case:

$ ./gogh o "" <input>

Answer 12

CoffeeScript/LiveScript/etc. (Node.js), 31 bytes

I'm not sure if any of those languages have been used yet, but here it is.

p=process;p.stdin.pipe p.stdout

Because it takes advantage of so few features, most of the CoffeeScript descendants with any notability would work with this code, including LiveScript, Coco, and IcedCoffeeScript.

Oh, and it does deal with infinite streams.

Answer 13

0815, 10 bytes

(interpreter, don't let the pirate icon scare you)

}: :!~$^: 

Note that there is a trailing space at the end. This one survives trimmers:

}:0:!~$^:0

They work the same.

Explanation

0815 has three memory registers: X, Y and Z. X is write-able, Z is read-able. Y cannot be directly accessed, but only with rotations. At start, X is 0x0, Y is 0x0 and Z is 0x0. 0815 only supports hexadecimal numbers. Labels point to a specific part in the code: the character after their definition. Here are the symbols used here:

• } : :: } defines a label, : starts its parameter (the label), is the label, and : closes the parameter, thus creating the label at this point (char 5).
• !: ! gets a byte from STDIN and stores it in X.
• ~ $: ~ is needed to rotate left, so that X is Y, Y is Z and Z is X. Then, $ is used to print the character in Z to the screen.
• ^ : : ^ jumps to the label specified if Z is not 0x0. Note that closing is not required on this step, as this is the end of the program. This allows an un-copiable null byte to be printed.

Answer 14

Seed, 7 Bytes

2 20093

Seed is a language which generates Befunge-93 with a length and a random seed. 2 is the length and 20093 is the random seed.

Answer 15

Gaot++, 113 bytes

bleeeet bleeeeeet bleeeeeeeeeeeet bleeeet b bleeeeeeeeeeeeet bleeeet b bleeeeeeet bleeeet bleeeeeeeeeet bleeeeeet

In Compressed Gaot++ it's 25 bytes:

4e6e12e4eb13e4eb7e4e10e6e

Cannot terminate on the offline interpreter, EOF is not recognized by the language :(

Online interpreter for both normal and compressed versions.

Answer 16

D, 65 71 68 bytes

import std.c.stdio;void main(){for(int c;c=~getchar,c;)putchar(~c);}

The original version couldn't handle infinitely long lines, but we should still do better than the existing D solution's 94 bytes, so here's the D version of Dennis' C solution.

std.c.stdio is deprecated in favor of core.stdc.stdio, as is this use of the comma operator, but both are still legal as of the current version.

Edit 2: Saved 3 bytes thanks to Zachary's suggestions

Answer 17

Casio FX-7000G, 3 bytes

Due to the low amount of storage, the calculator's programming mode used a tokenized language to save space. This makes it somewhat competitive for code-golf...

Program:

?→X

Explanation:

?       # Take input from user
 →X     # Store in variable X
        # Implicit: Print last value

Note that although this is 5 UTF-8 bytes, the calculator uses its own encoding to store this in 3.

Answer 18

NewbieFuck, 4 bytes

[.,]

I guess this is non-competing I'm not aware of any working interpreter, but I found the spec online (esolangs.org).

If I understand correctly, this is the same as Brainfuck, but [ ... ] is actually do-while loop (always executes the first time).

So, although the Brainfuck answer needs to take input before entering the loop, ,[.,], this language can omit the first input and use [.,]

Note: as end of input is signified by 0 in brainfuck, this cannot handle null bytes. It will also prepend a null byte to the output.

Answer 19

Perl, 13 or 14 bytes, depending on whether -e counts

perl -pe'BEGIN{$/=\1}'

Although perl's -p switch normally reads a line at a time, and thus would require the input data either be finite, or have newlines in it, we can change the input record separator $/.

If $/ is reference to a number, it causes perl to read fixed length records.

This has to happen inside a BEGIN block, or else at least one newline would be required in the input.

Answer 20

Lean Mean Bean Machine, 7 bytes

O
i
!
~

Marble spawns at O, i reads single character from STDIN to the marble's value, ! print's the marble's value as a Unicode character, ~ teleports the marble back up to the top.

Answer 21

Zucchini, 74 bytes

=0,1.12./.2.7./.3.6-/.4./.5./.1./.9,10.11+/.6.8./.9./.15./.13,/.13+/.12.0,

Answer 22

Evil, 4 bytes

mrwb

This language can only handle 1 character of input at a time.

m: Serves as a marking character for b.
r: Reads a single character from stdin and stores it in the accumulator.
w: Writes the value of the accumulator to stdout.
b: Searches backwards for m and continues execution from there.

(My) interpreter in C++ here: https://github.com/xprogram/esolot/tree/master/lang/evil

Original interpreter in Java here: http://web.archive.org/web/20070906133127/http://www1.pacific.edu/~twrensch/evil/evil.java

Answer 23

O, 7 bytes

1{io1}w

O is a new stack-based golfing language that aims to have as many 1-byte commands as possible. You can find O on GitHub.

1: pushes 1 for a truthy value.

{: begin code block (like ruby)

i: pushes input to stack

o: outputs top of stack

1: as above

}: ends code block

w: repeatedly does code block immediately before it while top of stack is truthy (pops top of stack)

Answer 24

Excel VBA, 6+1 = 7 Bytes

Since Excel VBA has no STDIN that handles input streams, the simplest way to address the prompt is to take input from a range on the ActiveSheet object (STDIN) and pass it onto the VBE immediate window (STDOUT)

?[A1];

Where [A1].Text has had ' preppended to its value (+1 Byte)

Alternative Version, 64 Bytes

To avoid the need for prepending the input with ', the user may instead check for an error or formula in [A1] and if so then return [A1].Formula else [A1] to the STDOUT in the VBE immediate window.

[B1]="=IsFormula(A1)":?IIf([IsErr(A1)]Or[B1],[A1].Formula,[A1]);

Answer 25

Samau, 0 bytes

An empty program in Samau is a cat program. It simply pushes the input onto the stack as a string, and prints the top of the stack.

Since Samau is written in Haskell, it can handle infinite streams.

Answer 26

Malbolge, 9,502 bytes

Full code here.

Only included here for the sake of completeness, not my code. Here is the author's page.

Answer 27

Wumpus, 6 bytes

i=)
o%

Try it online!

Explanation

First, to understand the control flow, we'll need to look at the actual program grid. Wumpus uses a triangular grid, so characters are alternatingly placed in upward and downward triangles. Furthermore, as opposed to many other Fungeoids, Wumpus does not use a wrapping grid. Instead, the instruction pointer (IP) is reflected off the boundary of the grid. That leads to the following flow through the grid:

Hence, Wumpus simply loops through the entire code (going right to left on the second line), i.e.

i=)%o

A single loop iteration is simple enough:

i   Read one byte N from STDIN and push it to the stack. EOF gives -1.
=   Duplicate.
)   Increment. Gives 0 at EOF.
%   Compute N % (N+1). At EOF this ends the program due to the division by zero.
    Otherwise, this just results in N itself.
o   Print the byte to STDOUT.

Answer 28

Cascade, 7 6 bytes

?.
;,^

Basically works by going in a loop outputting an input character until EOF is reached

I didn't realise the starting point wasn't required, -1 byte

Try it online!

Answer 29

Starry, 10 8 bytes

` , + .'

Try it Online!

Answer 30

1+, 13 bytes

1##,";1+1<1+#

Assuming EOF returns 0.

If error-terminating is allowed, it becomes fairly trivial:

1+, 7 bytes

1##,;1#