while (($_ = $*IN.get) != 42) { say $_; }

print while($_=<>)!=42 Any golfers want to show me how it's really done?

with Ada.Text_IO; use Ada.Text_IO; with Ada.Integer_Text_IO; use Ada.Integer_Text_IO; procedure test is x : integer; begin loop get(x); if x/=42 then put(x); new_line(1); else … read more

+[>>---------- [++++++++++<,----------] >-------------------------------------------------- >---------------------------------------------------- > [ <++++++++++++++++++++++++++++++++++++++++++++++++++++ <++++++++++++++… read more

{ if($0 != "42") print $0 else exit } And another alternative solution: /42/{exit} {print $0}

<?php $stdin = fopen('php://stdin', 'r'); while (true) { $input = fgets($stdin, 3); if ($input == 42) { break; } else { echo $input; } } fclose($stdin); ?>

while (l = gets.chomp.to_i) != 42 puts l end

I've noticed an influx of C# coders around here. I don't use the language much anymore for the more difficult problems because the I/O was so slow. The most difficult part to get around was that the input wasn't always… read more

ideone.com/FfZ2v package main import ( "fmt" ) func main() { number:=0 fmt.Scanln(&number) for number!= 42 { fmt.Println(number) fmt.Scanln(&number) } }

while {1} { gets stdin x if {$x == 42} break puts "$x\n" } exit 0

using System; public class Test { public static void Main() { int n; while ((n = int.Parse(Console.ReadLine()))!=42) Console.WriteLine(n); } }

int main() { while (sscanf(Stdio.stdin->gets(), "%d", int n), n!=42 && write(n+"\n")); return 0; }

PLEASE DO ,1 <- #1 PLEASE DO .4 <- #0 PLEASE DO .5 <- #0 PLEASE DO .99 <- #0 DO COME FROM (30) DO COME FROM (31) DO WRITE IN ,1 DO .1 <- ,1SUB#1 DO .2 <- .4 DO (1000) NEXT DO .4 <- .3~#255 DO (10) NEXT (42) … read more

section .bss buffer resb 1 section .text global _start _start: call ReadChar cmp al, '4' jne wait_enter ; we've got '4' … read more

while read a; do if [ $a -eq 42 ]; then exit else echo $a fi done

|c number| [ number:=0. [ (c := stdin next) asciiValue ~= 10 ] whileTrue: [number := (number * 10) + (c asciiValue) - 48.]. number ~= 42 ] whileTrue: [Transcript show: number printString; cr.] !

program TEST integer ans do read (*,*) ans if (ans.eq.42) stop write (*,*) ans enddo … read more

Only white spaces are relevant in the code below. [wink] CodeStartsHere:label 2 push 0 read push 0 retrieve push 42 sub dup jz 1 push 42 add print pus… read more

procedure main () while (l := read()) ~= 42 do write(l); end

Discussion on other languages and problems with all other SPOJ compilers