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 fulfil 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.

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

Dis, 2 bytes.

}{

Olmstead's cat program is too long, and it does not do really use the features of the language; just use that non-commands are NOP and the program counter loops over! Though the output would be slow, but this is pretty simple.

Try it online!

Answer 2

Java, 91 bytes

interface M{static void main(String[]a)throws Exception{System.in.transferTo(System.out);}}

Try it online!

Java, 101 bytes

interface M{static void main(String[]a)throws Exception{System.out.write(System.in.readAllBytes());}}

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Java, 119 bytes

interface M{static void main(String[]a)throws Throwable{for(int i;(i=System.in.read())>-1;)System.out.print((char)i);}}

Try it online!

Java, 120 bytes

interface M{static void main(String[]a){System.out.print(new java.util.Scanner(System.in).useDelimiter("\\A").next());}}

Try it online!

Answer 3

Mascarpone, 10 bytes

[,.:!]v*:!

Try It Online!

Answer 4

CSASM v2.5.1, 36 bytes

func main:
ink ""
print
call main
ret
end

Waits for a key to be pressed, then prints that key (the input key is also printed by ink due to how it's coded) forever.

Explanation:

ink prompts the user with whatever string is used as its operand -- an empty string in this case -- then gets the input from the user and pushes it to the stack as a char value.

print prints the str representation of whatever value is on the stack and does not append a newline.

The code will then call main, causing infinite recursion. Eventually, it will stop due to a StackOverflowException but it will work just fine up until that point. The program can be halted by doing Ctrl + X.

Answer 5

CLC-INTERCAL, 82 bytes.

DO;2<-#1DOCOMEFROM#9DOWRITEIN;2(1)DO:1<-;2SUB#1DDOCOMEFROM':1~:1'~#1(9)DOREADOUT;2

Copy and paste to try it online!

How it works

DONOTE it shall have at most one byte
DO;2<-#1

DOCOMEFROM#9
   DONOTE similar to getchar() but not really
   DOWRITEIN;2
   DONOTE checks if EOF
   (1)DO:1<-;2SUB#1
   DONOTE error exit
   D
   DONOTE come here unless EOF
   DOCOMEFROM':1~:1'~#1
   DONOTE no need to modify the array before outputting
   (9)DOREADOUT;2

How I/O works in CLC-INTERCAL

• Has four methods but I know only three.
• Numeric I/O. Applied for spot and two-spot variables, almost as in INTERCAL-72 and C-INTERCAL.
  • Input is done for each line. Each digit spelled. E.g. ONE THREE TWO.
  • Output is done for each line, in Roman number. Instead of overlines, backslashes are used to indicate "times 1M".
• Baudot I/O. Applied for tail. Input is done like gets(); outputs like puts() so it's done every line.
  • It uses extended Baudot character set; a character ` is missing.
• Binary I/O. Applied for hybrid array.
  • How WRITE IN works: if an array to be WRITTEN has N items, then N bytes are WRITTEN.
    1. Let io=172 and i=1.
    2. If i>N; then end.
    3. Let c=getchar().
    4. If c is EOF, then i-th item will be #0.
    5. If c is not EOF, then i-th item will be interleave(select(c,io),select(~c,~io)). In this expression c and io have 8 bits.
    6. If c is not EOF, then io=c.
    7. If c is not EOF, then most significant 16 bits of i-th item will be randomized so that at least one bit of them will be 1.
    8. Increment i. Go to step 2.
  • READING OUT is done in reverse way.
  • Variable "io" is initialized to 172 every time you WRITEIN and READOUT.

Answer 6

Python, 14 bytes

print(input())

Print input... couldn't be more self-explanatory.

Answer 7

makina, 6 3 bytes

I misread the requirements as needing a loop, whoops

Because makina is written in Python and uses Python's I/O methods, the information from the Python answer also applies to this answer.

P
i

Answer 8

C#, 76 bytes

for(int c=Console.Read();c!=-1;c=Console.Read())Console.Write((char)c);

Answer 9

C (clang), 37 bytes

main(c){for(;read(0,&c,1);)write(1);}

Stack reutilization inspired by this tip.

Try it online!

Answer 10

Trianguish, 8 bytes

00000000: 4041 01a1 040c 50bf

Try it online!

Answer 11

x86 assembly (32-bit Linux), 19 bytes

Turned out almost identical to Doorknob's answer, so I'm not sure where the one byte saved came from. Uses same tricks: registers start zeroed and syscalls don't clobber any registers besides eax. (Probably) segfaults on EOF.

global _start

section .text
_start:
        inc     edx         ; set edx = 1
        mov     ecx, esp    ; read/write to sp
loop:
        mov     al, 3       ; read call number 3    
                            ; ebx: read from stdin (fd=0)
                            ; edx: read 1 char
        int     0x80        ; syscall

        test    eax, eax    ; check if byte read
        jz      exit 

        mov     al, 4       ; write call number 
        inc     ebx         ; write to stdout (fd=1)
                            ; ecx: use char in buf
                            ; edx: write 1 char
        int     0x80        ; syscall

        dec     ebx         ; ebx = 0
        jmp     loop

exit:
                            ; segfault instead of exiting

Objdump:

08049000 <_start>:
 8049000:   42                      inc    edx
 8049001:   89 e1                   mov    ecx,esp

08049003 <loop>:
 8049003:   b0 03                   mov    al,0x3
 8049005:   cd 80                   int    0x80
 8049007:   85 c0                   test   eax,eax
 8049009:   74 08                   je     8049013 <exit>
 804900b:   b0 04                   mov    al,0x4
 804900d:   43                      inc    ebx
 804900e:   cd 80                   int    0x80
 8049010:   4b                      dec    ebx
 8049011:   eb f0                   jmp    8049003 <loop>

Answer 12

Trilangle, 14  8 bytes

<>i,@##o

Try it on the online interpreter!

The interpreter loads this into a triangular grid and adds trailing .s as necessary. When expanded, this program becomes:

   <
  > i
 , @ #
# o . .

The IP starts on the north corner headed southwest.

Starting on the red path, the interpreter hits the following instructions:

• <: Redirect the IP left, where it wraps around to the right side of the next row
• i: Read a single character from input
• >: Branch

When it hits the branch instruction, the IP changes direction depending on the sign of the character read. EOF is -1 and all other characters are positive, so it branches on EOF.

When it reads EOF, it continues on the green path, and quickly hits @, terminating the program. Otherwise, it continues on the blue path:

• o: Output a character
• ,: Pop from the stack
• #: Skip the next instruction (twice)

After this, it reaches the <, where it merges with the red path.

The , instruction is necessary to keep memory bounded when fed an infinite input stream (e.g. when piping yes into it).

Answer 13

Bitwise, 27 bytes

IN 1 &1 2
OUT 1 2
JMP &-3 2

Commented:

IN 1 &1 2   read a character from standard input into register 1 if literal 1 is nonzero, store result in 2
OUT 1 2     output the character in register 1 if register 2 is nonzero, discard result
JMP &-3 2   jump 3 (2) lines backwards if register 2 is nonzero

The proper program would look like this:

LABEL &1    create label 1
IN 1 &1 2   (same as previous)
OUT 1 2     (same as previous)
JMP @1 2    jump back to label 1 if register 2 is nonzero

Answer 14

shortC, 10 bytes

AW_=~G)P~_

Dennis' winning C answer in shortC. Try it online!

Answer 15

Pyt, 3 bytes

`ƥł

Try it online!

`...ł  loops while top of stack is not 0
ƥ      prints top element

Answer 16

Momema, 18 bytes

z1-9*0z00*-9z=+1*0

Try it online!

Explanation

                                                                  #  var a
z   1      #  label z0: jump past label z1                        #  loop body (initially skip this):
-9  *0     #            print chr [0]                             #    print chr a
z   0      #  label z1: jump past label z1 (no-op)                #  end loop body
0   *-9    #            [0] = read chr                            #    a = read chr
z   =+1*0  #  label z2: jump past label z((!!([0] + 1) + 2) % 3)  #  if a != -1, go to loop body

Answer 17

rk, 52 bytes

read: key #   
while # > -1 do
print: #
read: #
done

Requires the -e flag (remove necessity for rk:start). Try it online!

Ungolfed:

rk:start
  key #
  read: #

  while # > -1 do
    print: #
    read: #
  done
rk:end

Answer 18

Canvas, 0 bytes

Try it here!

Canvas automatically adds the input to the stack before doing anything else, and also automatically prints the object at the top of the stack once all other instructions have been completed. Therefore, an empty program functions as an implicit cat.

Answer 19

Implicit, 10 2 bytes

©"

© reads all input to the stack as ASCII characters, " turns the entire stack into a string.

Try it online!

A proper version

~.(-1@~.)&

Try it online! Explanation:

~+1(-1@;~+1)&
~            read character
 +1          increment
   (....     do
    -1        decrement
      @       print
       ~      read character
        +1    increment
          )  while top of stack truthy
           & exit (no implicit output)

All that incrementing and decrementing is to get the loop to exit on EOF (0).

Answer 20

Bitwise Cyclic Tag, 1 bit, 0.125 bytes

0

This will delete bits from the data tape (taking the starting data configuration as input) in order until there are none remaining to delete, at which point the program exits gracefully.

I'm on a phone now, but when I get the chance, I'll write an interpreter and post a TIO link.

Answer 21

Procedural Footnote Language, 33 bytes

[1]
[PFL1.0]
[1] [INPUT]
[PFLEND]

This program simply defines the body of the document as a reference to footnote [1], then defines the footnote as the value of the input from STDIN. The evaluated body of the document is printed as a result.

Answer 22

Reality, 0 bytes

This assumes that input is given (if it is not it will output something else)

Answer 23

Z80Golf, 7 bytes

00000000: cd03 8030 0176 d5                        ...0.v.

Try it online!

Disassembly

start:
  call $8003
  jr nc, skip
  halt
skip:
  push de

• call $8003: Calls getchar. It sets carry flag on EOF, otherwise writes a byte to register a.
• jr nc, skip: If carry flag is not set, jump to skip:. Otherwise the program hits halt and terminates.
• push de: Pushes two zero bytes to the stack. The program runs through the NOPs in memory until it hits putchar at address $8000. Then the value in the register a is printed, and the program returns to the popped address, which is zero (start of the program).

This happens to be a slightly modified version of the one on the Esolangs page, where the position of push de is different. Using bc or hl instead of de equally works.

Answer 24

SNUSP (Modular), 11 bytes

/$\
. ,
\?/

Try it online!

Please note that the TIO doesn't work with input correctly, but the program is correct. In SNUSP, the instruction pointer starts at the $ moving right, and bounces off of slashes. When the IP goes over a ,, it reads a character from STDIN to the current memory cell. The ? tests whether or not the current value is 0. If it is, then then IP jumps over the next instruction (the \) and goes off the end of the program, terminating. Otherwise, the IP goes onto the \, reflects, and hits the ., printing the current value as a character to STDOUT. When the IP goes over the $ again, it doesn't do anything.

Answer 25

bit, 16 bytes

IN
PRINT
PRINTLN

Explanation:

IN $$ Take input from user and push to stack
PRINT $$ add to printing queue
PRINTLN $$ Print the printing queue with a trailing newline

Bit is my own language, get the interpreter from GitHub and run:

java -jar bit.jar filename.bit

Answer 26

Backhand, 2 bytes

io

Try it online!

Seems obvious that i gets a character from the input and o outputs that character. However, the control flow is different from normal 2D languages in that it doesn't wrap around. Instead, the pointer moves three spaces, reflecting off the sides twice for each instruction.

Answer 27

Dots, 43453 bytes

....................................................................................................................................................................................................................................................................................................................................................................... and so on until 43453 bytes

I'll golf this later Nevermind, this is the least possible number of bytes, unless we assume that STDIN is one character long

Answer 28

Common Lisp, 46 bytes

This is another, shorte, solution:

(loop while(setq x(read-char t()))do(princ x))

Try it online!

Note in that Common Lisp every output on standard output always produces a newline at end (due to its nature as interactive language). If the program above is run on a different output stream, the file does not contains the final newline if this is not present in the input stream.

Answer 29

Re:direction -A, 1 byte of codepage 437

◄

Try it online!

The -A option allows Re:direction to interpret the input as the list of its character codes, rather than trying to parse numbers from it.

Explanation

In Re:direction, an arrow on a line by itself is a halt command. Arrows also change the command queue, but leftwards and upwards arrows don't have an effect on how the command queue is interpreted as I/O.

The input specifies the initial command queue, and the output is taken from the final command queue. So if we don't change it at all before halting (or change it only by adding leftwards or upwards arrows), the output will be the same as the input.

Re:direction also supports a more efficient character encoding than codepage 437, but the program wouldn't be any shorter.

Answer 30

!@#$%^&*()_+, 13 bytes

*!^(!_++@*!^)

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Explanation

*!^(!_++@*!^)
*                add input to top of stack
 !               duplicate
  ^              increment
   (        )    while not zero (while not EOF):
    !_++         pop incremented entry (duplicate, negate, add twice)
        @        output original input
         *       add input to top of stack
          !      duplicate
           ^     increment