Binary digits

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Task
Binary digits
You are encouraged to solve this task according to the task description, using any language you may know.

The task is to output the sequence of binary digits for a given non-negative integer.

The decimal value 5, should produce an output of 101 The decimal value 50 should produce an output of 110010 The decimal value 9000 should produce an output of 10001100101000

The results can be achieved using builtin radix functions within the language, if these are available, or alternatively a user defined function can be used. The output produced should consist just of the binary digits of each number followed by a newline. There should be no other whitespace, radix or sign markers in the produced output, and leading zeros should not appear in the results.

Contents

[edit] 0815

}:r:|~    Read numbers in a loop.
  }:b:    Treat the queue as a stack and
    <:2:= accumulate the binary digits
    /=>&~ of the given number.
  ^:b:
  <:0:->  Enqueue negative 1 as a sentinel.
  {       Dequeue the first binary digit.
  }:p:
    ~%={+ Rotate each binary digit into place and print it.
  ^:p:
  <:a:~$  Output a newline.
^:r:
Output:

Note that 0815 reads numeric input in hexadecimal. 

echo -e "5\n32\n2329" | 0815 bin.0
101
110010
10001100101001

[edit] ACL2

(include-book "arithmetic-3/top" :dir :system)
 
(defun bin-string-r (x)
   (if (zp x)
       ""
       (string-append
        (bin-string-r (floor x 2))
        (if (= 1 (mod x 2))
            "1"
            "0"))))
 
(defun bin-string (x)
   (if (zp x)
       "0"
       (bin-string-r x)))

[edit] Ada

with Ada.Text_IO;
 
procedure Binary_Output is
 
   package IIO is new Ada.Text_IO.Integer_IO(Integer);
 
   function To_Binary(N: Natural) return String is
      S: String(1 .. 1000); -- more than plenty!
      Left:  Positive := S'First;
      Right: Positive := S'Last;
   begin
      IIO.Put(To => S, Item => N, Base => 2); -- This is the conversion!
      -- Now S is a String with many spaces and some "2#...#" somewhere.
      -- We only need the "..." part without spaces or base markers.
      while S(Left) /= '#' loop
         Left := Left + 1;
      end loop;
      while S(Right) /= '#' loop
         Right := Right - 1;
      end loop;
      return S(Left+1 .. Right-1);
   end To_Binary;
 
begin
   Ada.Text_IO.Put_Line(To_Binary(5));    -- 101
   Ada.Text_IO.Put_Line(To_Binary(50));   -- 110010
   Ada.Text_IO.Put_Line(To_Binary(9000)); -- 10001100101000
end Binary_Output;

[edit] Aime

o_xinteger(2, 0);
o_byte('\n');
o_xinteger(2, 5);
o_byte('\n');
o_xinteger(2, 50);
o_byte('\n');
o_xinteger(2, 9000);
o_byte('\n');
Output:
0
101
110010
10001100101000

[edit] ALGOL 68

Works with: ALGOL 68 version Revision 1.
Works with: ALGOL 68G version Any - tested with release algol68g-2.3.3.
File: Binary_digits.a68
#!/usr/local/bin/a68g --script #
 
printf((
  $g" => "2r3d l$, 5, BIN 5,
  $g" => "2r6d l$, 50, BIN 50,
  $g" => "2r14d l$, 9000, BIN 9000
));
 
# or coerce to an array of BOOL #
print((
  5, " => ", []BOOL(BIN 5)[bits width-3+1:], new line,
  50, " => ", []BOOL(BIN 50)[bits width-6+1:], new line,
  9000, " => ", []BOOL(BIN 9000)[bits width-14+1:], new line
))
Output
         +5 => 101
        +50 => 110010
      +9000 => 10001100101000
         +5 => TFT
        +50 => TTFFTF
      +9000 => TFFFTTFFTFTFFF

[edit] AutoHotkey

MsgBox % NumberToBinary(5) ;101
MsgBox % NumberToBinary(50) ;110010
MsgBox % NumberToBinary(9000) ;10001100101000
 
NumberToBinary(InputNumber)
{
 While, InputNumber
  Result := (InputNumber & 1) . Result, InputNumber >>= 1
 Return, Result
}

[edit] AutoIt

 
ConsoleWrite(IntToBin(50) & @CRLF)
 
Func IntToBin($iInt)
	$Stack = ObjCreate("System.Collections.Stack")
	Local $b = -1, $r = ""
	While $iInt <> 0
		$b = Mod($iInt, 2)
		$iInt = INT($iInt/2)
		$Stack.Push ($b)
	WEnd
	For $i = 1 TO $Stack.Count
		$r &= $Stack.Pop
	Next
	Return $r
EndFunc   ;==>IntToBin
 

[edit] AWK

BEGIN {
  print tobinary(5)
  print tobinary(50)
  print tobinary(9000)
}
 
function tobinary(num) {
  outstr = ""
  l = num
  while ( l ) {
    if ( l%2 == 0 ) {
      outstr = "0" outstr
    } else {
      outstr = "1" outstr
    }
    l = int(l/2)
  }
  # Make sure we output a zero for a value of zero
  if ( outstr == "" ) {
    outstr = "0"
  }
  return outstr
}

[edit] Batch File

This num2bin.bat file handles non-negative input as per the requirements with no leading zeros in the output. Batch only supports signed integers. This script also handles negative values by printing the appropriate two's complement notation. 

@echo off
:num2bin    IntVal [RtnVar]
  setlocal enableDelayedExpansion
  set /a n=%~1
  set rtn=
  for /l %%b in (0,1,31) do (
    set /a "d=n&1, n>>=1"
    set rtn=!d!!rtn!
  )
  for /f "tokens=* delims=0" %%a in ("!rtn!") do set rtn=%%a
  (endlocal & rem -- return values
    if "%~2" neq "" (set %~2=%rtn%) else echo %rtn%
  )
exit /b

[edit] BBC BASIC

      FOR num% = 0 TO 16
        PRINT FN_tobase(num%, 2, 0)
      NEXT
      END
 
      REM Convert N% to string in base B% with minimum M% digits:
      DEF FN_tobase(N%,B%,M%)
      LOCAL D%,A$
      REPEAT
        D% = N%MODB%
        N% DIV= B%
        IF D%<0 D% += B%:N% -= 1
        A$ = CHR$(48 + D% - 7*(D%>9)) + A$
        M% -= 1
      UNTIL (N%=FALSE OR N%=TRUE) AND M%<=0
      =A$

The above is a generic "Convert to any base" program. Here is a faster "Convert to Binary" program: 

PRINT FNbinary(5)
PRINT FNbinary(50)
PRINT FNbinary(9000)
END
 
DEF FNbinary(N%)
LOCAL A$
REPEAT
  A$ = STR$(N% AND 1) + A$
  N% = N% >>> 1  : REM BBC Basic prior to V5 can use N% = N% DIV 2
UNTIL N% = 0
=A$

[edit] bc

Translation of: dc
obase = 2
5
50
9000
quit

[edit] Bracmat

  ( dec2bin
  =   bit bits
    .   :?bits
      &   whl
        ' ( !arg:>0
          & mod$(!arg,2):?bit
          & div$(!arg,2):?arg
          & !bit !bits:?bits
          )
      & (str$!bits:~|0)
  )
& 0 5 50 9000 423785674235000123456789:?numbers
&   whl
  ' ( !numbers:%?dec ?numbers
    & put$(str$(!dec ":\n" dec2bin$!dec \n\n))
    )
;

Output: 

0:
0

5:
101

50:
110010

9000:
10001100101000

423785674235000123456789:
1011001101111010111011110101001101111000000000000110001100000100111110100010101

[edit] Brainf***

This is almost an exact duplicate of Count in octal#Brainf***. It outputs binary numbers until it is forced to terminate or the counter overflows to 0. 

+[            Start with n=1 to kick off the loop
[>>++<<       Set up {n 0 2} for divmod magic
[->+>-        Then
[>+>>]>       do
[+[-<+>]>+>>] the
<<<<<<]       magic
>>>+          Increment n % 2 so that 0s don't break things
>]            Move into n / 2 and divmod that unless it's 0
-<            Set up sentinel ‑1 then move into the first binary digit
[++++++++ ++++++++ ++++++++ Add 47 to get it to ASCII
 ++++++++ ++++++++ +++++++. and print it
[<]<]         Get to a 0; the cell to the left is the next binary digit
>>[<+>-]      Tape is {0 n}; make it {n 0}
>[>+]         Get to the ‑1
<[[-]<]       Zero the tape for the next iteration
++++++++++.   Print a newline
[-]<+]        Zero it then increment n and go again

[edit] Burlesque

 
blsq ) {5 50 9000}{2B!}m[uN
101
110010
10001100101000
 

[edit] C

Converts int to a string. 

#include <stdio.h>
 
void bin(int x, char *s)
{
	char*_(int x){
		*(s = x ? _(x >> 1) : s) = (x & 1) + '0';
		return ++s;
	}
	*_(x) = 0;
}
 
int main()
{
	char a[100];
	int i;
	for (i = 0; i <= 1984; i += 31)
		bin(i, a), printf("%4d: %s\n", i, a);
 
	return 0;
}

Converts int to a string. 

#include <stdio.h>
 
void IntToBitString(unsigned int number)
{
    int num_bits = sizeof(unsigned int) * 8;
 
    bool startPrinting = false;
    for (int bit_pos=num_bits-1; bit_pos >= 0; bit_pos--)
    {
        bool isBitSet = (number & (1<<bit_pos)) != 0;
 
        if (!startPrinting && isBitSet)
            startPrinting = true;
 
        if (startPrinting || bit_pos==0)
            printf("%s", isBitSet ? "1":"0");
    }
 
    printf("\r\n");
}
 
int main()
{
    IntToBitString(0);
    IntToBitString(5);
    IntToBitString(50);
    IntToBitString(9000);
 
    return 0;
}

[edit] C++

#include <iostream>
#include <bitset>
#include <string>
#include <climits>
 
void print_bin(int n)
{
    // convert to binary, then to string
    std::string str = std::bitset<CHAR_BIT * sizeof n>(n).to_string();
    // trim leading zeroes
    if(n == 0)
       str = "0";
    else
       str = str.substr(str.find('1'));
    // output
    std::cout << str << '\n';
}
 
int main()
{
    print_bin(0);
    print_bin(5);
    print_bin(50);
    print_bin(9000);
}

[edit] Clojure

(Integer/toBinaryString 5)
(Integer/toBinaryString 50)
(Integer/toBinaryString 9000)

[edit] COBOL

       IDENTIFICATION DIVISION.
       PROGRAM-ID. SAMPLE.
 
       DATA DIVISION.
       WORKING-STORAGE SECTION.
 
         01 binary_number   pic X(21).
         01 str             pic X(21).
         01 binary_digit    pic X.
         01 digit           pic 9.
         01 n               pic 9(7).
         01 nstr            pic X(7).
 
       PROCEDURE DIVISION.
         accept nstr
         move nstr to n
         perform until n equal 0
           divide n by 2 giving n remainder digit
           move digit to binary_digit
           string binary_digit  DELIMITED BY SIZE
                  binary_number DELIMITED BY SPACE
                  into str
           move str to binary_number
         end-perform.
         display binary_number
         stop run.
 

[edit] CoffeeScript

binary = (n) ->
  new Number(n).toString(2)
 
console.log binary n for n in [5, 50, 9000]

[edit] Common Lisp

Just print the number with "~b": 

(format t "~b" 5)

[edit] C#

using System;
 
class Program
{
    static void Main()
    {
        foreach (var number in new[] { 5, 50, 9000 })
        {
            Console.WriteLine(Convert.ToString(number, 2));
        }
    }
}

Output: 

101
110010
10001100101000

[edit] D

import std.stdio;
 
void main() {
    foreach (i; 0 .. 16)
        writefln("%b", i);
}
Output:
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111

[edit] Dart

String binary(int n) {
  if(n<0)
    throw new IllegalArgumentException("negative numbers require 2s complement");
  if(n==0) return "0";
  String res="";
  while(n>0) {
    res=(n%2).toString()+res;
    n=(n/2).toInt();
  }
  return res;
}
 
main() {
  print(binary(0));
  print(binary(1));
  print(binary(5));
  print(binary(10));
  print(binary(50));
  print(binary(9000));
  print(binary(65535));
  print(binary(0xaa5511ff));
  print(binary(0x123456789abcde));
  // fails due to precision limit
  print(binary(0x123456789abcdef));
}

[edit] dc

2o 5p 50p 9000p
Output:
101
110010
10001100101000

[edit] Delphi/Pascal

Library: SysUtils
procedure Binary_Digits;
 
  function IntToBinStr(AInt : integer) : string;
 
  begin
    Result := '';
    while AInt > 0 do
      begin
        Result := Chr(Ord('0')+(AInt mod 2))+Result;
        AInt := AInt div 2;
      end;
  end;
 
begin
  writeln('   5: '+IntToBinStr(5));
  writeln('  50: '+IntToBinStr(50));
  writeln('9000: '+IntToBinStr(9000));
end;
   5: 101
  50: 110010
9000: 10001100101000

[edit] Elena

#define std'dictionary'*.
#define ext'convertors'*.
 
// --- Program ---
 
#symbol Program =
[
    'program'output write &numeric:5 &radix:2 &:eintformatter << "%n".
    'program'output write &numeric:50 &radix:2 &:eintformatter << "%n".
    'program'output write &numeric:9000 &radix:2 &:eintformatter << "%n".
].

[edit] Erlang

lists:map( fun(N) -> io:fwrite("~.2B~n", [N]) end, [5, 50, 9000]).

Output: 

101
110010
10001100101000

[edit] Euphoria

function toBinary(integer i)
    sequence s
    s = {}
    while i do
        s = prepend(s, '0'+and_bits(i,1))
        i = floor(i/2)
    end while
    return s
end function
 
puts(1, toBinary(5) & '\n')
puts(1, toBinary(50) & '\n')
puts(1, toBinary(9000) & '\n')

[edit] F#

Translation of C# 

 
open System
for i in [5; 50; 9000] do (Console.WriteLine Convert.ToString (i, 2))
 

[edit] Factor

USING: io kernel math math.parser ;
 
5 >bin print
50 >bin print
9000 >bin print

[edit] Forth

9000 50 5
2 base !
. . .
decimal

[edit] Fortran

Please find compilation instructions and the example run at the start of the FORTRAN90 source that follows. Thank you. 

 
!-*- mode: compilation; default-directory: "/tmp/" -*-
!Compilation started at Sun May 19 23:14:14
!
!a=./F && make $a && $a < unixdict.txt
!f95 -Wall -ffree-form F.F -o F
!101
!110010
!10001100101000
!
!Compilation finished at Sun May 19 23:14:14
!
!
!   tobin=: -.&' '@":@#:
!   tobin 5
!101
!   tobin 50
!110010
!   tobin 9000
!10001100101000
 
program bits
  implicit none
  integer, dimension(3) :: a
  integer :: i
  data a/5,50,9000/
  do i = 1, 3
    call s(a(i))
  enddo
 
contains
 
  subroutine s(a)
    integer, intent(in) :: a
    integer :: i
    if (a .eq. 0) then
      write(6,'(a)')'0'
      return
    endif
    do i = 31, 0, -1
      if (btest(a, i)) exit
    enddo
    do while (0 .lt. i)
      if (btest(a, i)) then
        write(6,'(a)',advance='no')'1'
      else
        write(6,'(a)',advance='no')'0'
      endif
      i = i-1
    enddo
    if (btest(a, i)) then
      write(6,'(a)')'1'
    else
      write(6,'(a)')'0'
    endif
  end subroutine s
 
end program bits
 

[edit] Go

package main
 
import (
	"fmt"
)
 
func main() {
	for i := 0; i < 16; i++ {
		fmt.Printf("%b\n", i)
	}
}
Output:
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111

[edit] Groovy

Solutions: 

print '''
  n        binary
----- ---------------
'''
[5, 50, 9000].each {
    printf('%5d %15s\n', it, Integer.toBinaryString(it))
}

Output: 

  n        binary
----- ---------------
    5             101
   50          110010
 9000  10001100101000

[edit] Haskell

import Data.List
import Numeric
import Text.Printf
 
-- Use the built-in function showIntAtBase.
toBin n = showIntAtBase 2 ("01" !!) n ""
 
-- Implement our own version.
toBin' 0 = []
toBin' x =  (toBin' $ x `div` 2) ++ (show $ x `mod` 2)
 
printToBin n = putStrLn $ printf "%4d  %14s  %14s" n (toBin n) (toBin' n)
 
main = do
  putStrLn $ printf "%4s  %14s  %14s" "N" "toBin" "toBin'"
  mapM_ printToBin [5, 50, 9000]
Sample output:
   N           toBin          toBin'
   5             101             101
  50          110010          110010
9000  10001100101000  10001100101000

[edit] Icon and Unicon

There is no built-in way to output the bit string representation of an whole number in Icon and Unicon. There are generalized radix conversion routines in the Icon Programming Library that comes with every distribution. This procedure is a customized conversion routine that will populate and use a tunable cache as it goes. 

procedure main()
every i := 5 | 50 | 255 | 1285 | 9000 do
  write(i," = ",binary(i))
end
 
procedure binary(n)                      #: return bitstring for integer n
static CT, cm, cb
initial {
   CT := table()                         # cache table for results
   cm := 2 ^ (cb := 4)                   # (tunable) cache modulus & pad bits 
   }   
 
b := ""                                  # build reversed bit string
while n > 0 do {                         # use cached result ...
   if not (b ||:= \CT[1(i := n % cm, n /:= cm) ]) then {                     
      CT[j := i] := ""                   # ...or start new cache entry
      while j > 0 do 
         CT[i] ||:=  "01"[ 1(1+j % 2, j /:= 2 )]
      b ||:= CT[i] := left(CT[i],cb,"0") # finish cache with padding
      }
   }
return reverse(trim(b,"0"))              # nothing extraneous
end

Output: 

5 = 101
50 = 110010
255 = 11111111
1285 = 10100000101
9000 = 10001100101000

[edit] J

   tobin=: -.&' '@":@#:
   tobin 5
101
   tobin 50
110010
   tobin 9000
10001100101000

Algorithm: Remove spaces from the character list which results from formatting the binary list which represents the numeric argument.

I am using implicit output.

[edit] Java

public class Main {
    public static void main(String[] args) {
        System.out.println(Integer.toBinaryString(5));
        System.out.println(Integer.toBinaryString(50));
        System.out.println(Integer.toBinaryString(9000));
    }
}

Output: 

101
110010
10001100101000

[edit] JavaScript

function toBinary(number) {
  return new Number(number).toString(2);
}
var demoValues = [5, 50, 9000];
for (var i=0; i<demoValues.length; ++i) {
  print(toBinary(demoValues[i])); // alert() in a browser, wscript.echo in WSH, etc.
}

Output: 

101
110010
10001100101000

[edit] Joy

HIDE
_ == [null] [pop] [2 div swap] [48 + putch] linrec
IN
int2bin == [null] [48 + putch] [_] ifte '\n putch
END

Using int2bin: 

0 setautoput
0 int2bin
5 int2bin
50 int2bin
9000 int2bin.

[edit] K

  tobin: ,/$2_vs
  tobin' 5 50 9000
("101"
 "110010"
 "10001100101000")

[edit] Lang5

'%b '__number_format set
[5 50 9000] [3 1] reshape .
Output:
[
  [ 101  ]
  [ 110010  ]
  [ 10001100101000  ]
]

[edit] Liberty BASIC

for a = 0 to 16
print a;"=";dec2bin$(a)
next
a=50:print a;"=";dec2bin$(a)
a=254:print a;"=";dec2bin$(a)
a=9000:print a;"=";dec2bin$(a)
wait
 
function dec2bin$(num)
   if num=0 then dec2bin$="0":exit function
    while num>0
        dec2bin$=str$(num mod 2)+dec2bin$
        num=int(num/2)
    wend
end function
 

[edit] Locomotive Basic

10 PRINT BIN$(5)
20 PRINT BIN$(50)
30 PRINT BIN$(9000)

Output: 

101
110010
10001100101000

[edit] LOLCODE

This example is incorrect. The task is to convert numbers into binary, not count in binary. Please fix the code and remove this message.

This program prints binary digits until it is forced to terminate or the counter overflows to 0. It's almost an exact duplicate of Count in octal#LOLCODE. 

HAI 1.3
 
HOW IZ I binary YR num
    I HAS A digit, I HAS A bin ITZ ""
    IM IN YR binarizer
        digit R MOD OF num AN 2
        bin R SMOOSH digit bin MKAY
        num R QUOSHUNT OF num AN 2
        NOT num, O RLY?
            YA RLY, FOUND YR bin
        OIC
    IM OUTTA YR binarizer
IF U SAY SO
 
IM IN YR printer UPPIN YR num
    VISIBLE I IZ binary YR num MKAY
IM OUTTA YR printer
 
KTHXBYE

[edit] Maple

 
> printf( "%d\n", convert( 50, 'binary' ) );
110010
> printf( "%d\n", convert( 9000, 'binary' ) );
10001100101000
 

[edit] Mathematica

StringJoin @@ ToString /@ IntegerDigits[50, 2]

[edit] MATLAB / Octave

  dec2bin(5)
  dec2bin(50)
  dec2bin(9000)

The output is a string containing ascii(48) (i.e. '0') and ascii(49) (i.e. '1').

[edit] Maxima

digits([arg]) := block(
   [n: first(arg), b: if length(arg) > 1 then second(arg) else 10, v: [ ], q],
   if n = 0 then [0] else (
      while n # 0 do (
         [n, q]: divide(n, b),
         v: endcons(q, v)
      ),
      v
   )
)$
 
binary(n) := simplode(reverse(digits(9000, 2)))$
 
binary(9000);
"10001100101000"

[edit] Mercury

:- module binary_digits.
:- interface.
 
:- import_module io.
:- pred main(io::di, io::uo) is det.
 
:- implementation.
:- import_module int, list, string.
 
main(!IO) :-
    list.foldl(print_binary_digits, [5, 50, 9000], !IO).
 
:- pred print_binary_digits(int::in, io::di, io::uo) is det.
 
print_binary_digits(N, !IO) :-
    io.write_string(int_to_base_string(N, 2), !IO),
    io.nl(!IO).

[edit] Modula-3

MODULE Binary EXPORTS Main;
 
IMPORT IO, Fmt;
 
VAR num := 10;
 
BEGIN
  IO.Put(Fmt.Int(num, 2) & "\n");
  num := 150;
  IO.Put(Fmt.Int(num, 2) & "\n");
END Binary.

Output: 

1010
10010110

[edit] NetRexx

/* NetRexx */
options replace format comments java crossref symbols nobinary
 
runSample(arg)
return
 
method getBinaryDigits(nr) public static
  bd = nr.d2x.x2b.strip('L', 0)
  if bd.length = 0 then bd = 0
  return bd
 
method runSample(arg) public static
  parse arg list
  if list = '' then list = '0 5 50 9000'
  loop n_ = 1 to list.words
    w_ = list.word(n_)
    say w_.right(20)':' getBinaryDigits(w_)
    end n_
Output:
                   0: 0
                   5: 101
                  50: 110010
                9000: 10001100101000

[edit] Objeck

 
bundle Default {
  class Binary {
    function : Main(args : String[]) ~ Nil {
      5->ToBinaryString()->PrintLine();
      50->ToBinaryString()->PrintLine();
      9000->ToBinaryString()->PrintLine();
    }
  }
}
 

[edit] OCaml

let bin_of_int d =
  if d < 0 then invalid_arg "bin_of_int" else
  if d = 0 then "0" else
  let rec aux acc d =
    if d = 0 then acc else
    aux (string_of_int (d land 1) :: acc) (d lsr 1)
  in
  String.concat "" (aux [] d)
 
let () =
  let d = read_int () in
  Printf.printf "%8s\n" (bin_of_int d)

[edit] OxygenBasic

The Assembly code uses block structures to minimise the use of labels. 

 
 
function BinaryBits(sys n) as string
  string buf=nuls 32
  sys p=strptr buf
  sys le
  mov eax,n
  mov edi,p
  mov ecx,32
  '
  'STRIP LEADING ZEROS
  (
   dec ecx
   jl fwd done
   shl eax,1
   jnc repeat
  )
  'PLACE DIGITS
  '
  mov byte [edi],49 '1'
  inc edi
  (
   cmp ecx,0
   jle exit
   mov dl,48 '0'
   shl eax,1
   (
    jnc exit
    mov dl,49 '1'
   )
   mov [edi],dl
   inc edi
   dec ecx
   repeat
  )
  done:
  '
  sub edi,p
  mov le,edi
  if le then return left buf,le
  return "0"
end function
 
print BinaryBits 0xaa 'result 10101010
 

[edit] PARI/GP

bin(n:int)=concat(apply(s->Str(s),binary(n)))

[edit] Perl

for (5, 50, 9000) {
  printf "%b\n", $_;
}
101
110010
10001100101000

[edit] Perl 6

say .fmt("%b") for 5, 50, 9000;
101
110010
10001100101000

[edit] PHP

<?php
echo decbin(5);
echo decbin(50);
echo decbin(9000);

Output: 

101
110010
10001100101000

[edit] PicoLisp

: (bin 5)
-> "101"
 
: (bin 50)
-> "110010"
 
: (bin 9000)
-> "10001100101000"

[edit] PL/I

Displays binary output trivially, but with leading zeros: 

put list (25) (B);
Output: 0011001

With leading zero suppression: 

   declare text character (50) initial (' ');
 
   put string(text) edit (25) (b);
   put skip list (trim(text, '0'));
 
   put string(text) edit (2147483647) (b);
   put skip list (trim(text, '0'));

Output: 

11001
1111111111111111111111111111111

[edit] PowerShell

Library: Microsoft .NET Framework
@(5,50,900) | foreach-object { [Convert]::ToString($_,2) }

Output: 

101
110010
1110000100

[edit] Prolog

Works with: SWI Prolog
Works with: GNU Prolog
 
binary(X) :- format('~2r~n', [X]).
main :- maplist(binary, [5,50,9000]), halt.
 
Sample output: 
101
110010
10001100101000

[edit] PureBasic

If OpenConsole()
  PrintN(Bin(5))    ;101
  PrintN(Bin(50))   ;110010
  PrintN(Bin(9000)) ;10001100101000
 
  Print(#CRLF$ + #CRLF$ + "Press ENTER to exit"): Input()
  CloseConsole()
EndIf

Sample output: 

101
110010
10001100101000

[edit] Python

Works with: Python version 3.X and 2.6+
>>> for i in range(16): print('{0:b}'.format(i))
 
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
Works with: Python version 3.X and 2.6+
>>> for i in range(16): print(bin(i))
 
0b0
0b1
0b10
0b11
0b100
0b101
0b110
0b111
0b1000
0b1001
0b1010
0b1011
0b1100
0b1101
0b1110
0b1111

Pre-Python 2.6: 

>>> oct2bin = {'0': '000', '1': '001', '2': '010', '3': '011', '4': '100', '5': '101', '6': '110', '7': '111'}
>>> bin = lambda n: ''.join(oct2bin[octdigit] for octdigit in '%o' % n).lstrip('0') or '0'
>>> for i in range(16): print(bin(i))
 
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111

[edit] Racket

 
#lang racket
;; Option 1: binary formatter
(for ([i 16]) (printf "~b\n" i))
;; Option 2: explicit conversion
(for ([i 16]) (displayln (number->string i 2)))
 

[edit] Retro

9000 50 5  3 [ binary putn cr decimal ] times

[edit] REXX

This version handles the special case of zero more elegantly.

[edit] version 1

Note: some REXX interpreters have a D2B [Decimal to Binary] bif. 

/*REXX program demonstrates converting  decimal  ───►  binary.          */
numeric digits 1000
x.1 = 0
x.2 = 5
x.3 = 50
x.4 = 9000
              do j=1  for 4
              y = x2b(d2x(x.j)) + 0
              say  right(x.j,20)  'decimal, and in binary:'  y
              end   /*j*/
                                        /*stick a fork in it, we're done.*/

output 

                   0 decimal, and in binary: 0
                   5 decimal, and in binary: 101
                  50 decimal, and in binary: 110010
                9000 decimal, and in binary: 10001100101000

[edit] version 2

This version handles the case of zero as a special case. 

/*REXX program demonstrates converting  decimal  ───►  binary.          */
x.1 = 0
x.2 = 5
x.3 = 50
x.4 = 9000
              do j=1  for 4
              y = strip( x2b( d2x( x.j )), 'L', 0)
              if y==''  then y=0
              say  right(x.j,20)  'decimal, and in binary:'  y
              end   /*j*/
                                        /*stick a fork in it, we're done.*/

output is identical to the 1st version.

[edit] version 3

This version handles the case of zero a bit more obtusely, but concisely. 

/*REXX program demonstrates converting  decimal  ───►  binary.          */
x.1 = 0
x.2 = 5
x.3 = 50
x.4 = 9000
              do j=1  for 4
              y = word( strip( x2b( d2x( x.j )), 'L', 0)     0, 1)
              say  right(x.j,20)  'decimal, and in binary:'  y
              end   /*j*/
                                       /*stick a fork in it, we're done.*/

output is identical to the 1st version.

[edit] Ruby

[5,50,9000].each do |n|
  puts "%b" % n
end

or 

for n in [5,50,9000]
  puts n.to_s(2)
end

Output: 

101
110010
10001100101000

[edit] Run BASIC

input "Number to convert:";a
while 2^(n+1) < a
 n = n + 1
wend
 
for i = n to 0 step -1
  x = 2^i
  if a >= x then 
    print 1;
    a = a - x
   else 
    print 0;
  end if
next

Output: 

Number to convert:?9000
10001100101000

[edit] Scala

5 toBinaryString     // 101
50 toBinaryString    // 110010
9000 toBinaryString  // 10001100101000

[edit] Scheme

(display (number->string 5 2)) (newline)
(display (number->string 50 2)) (newline)
(display (number->string 9000 2)) (newline)

[edit] Seed7

$ include "seed7_05.s7i";
 
const proc: main is func
  local
    var integer: number is 0;
  begin
    for number range 0 to 16 do
      writeln(str(number, 2));
    end for;
  end func;

Output: 

0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
10000

[edit] Smalltalk

5 printOn: Stdout radix:2
50 printOn: Stdout radix:2
9000 printOn: Stdout radix:2

or: 

#(5 50 9000) do:[:each | each printOn: Stdout radix:2. Stdout cr]

[edit] Standard ML

print (Int.fmt StringCvt.BIN 5 ^ "\n");
print (Int.fmt StringCvt.BIN 50 ^ "\n");
print (Int.fmt StringCvt.BIN 9000 ^ "\n");

[edit] Tcl

proc num2bin num {
    # Convert to _fixed width_ big-endian 32-bit binary
    binary scan [binary format "I" $num] "B*" binval
    # Strip useless leading zeros by reinterpreting as a big decimal integer
    scan $binval "%lld"
}

Demonstrating: 

for {set x 0} {$x < 16} {incr x} {
    puts [num2bin $x]
}
puts "--------------"
puts [num2bin 5]
puts [num2bin 50]
puts [num2bin 9000]

Output: 

0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
--------------
101
110010
10001100101000

[edit] TI-83 BASIC

Using Standard TI-83 BASIC 

PROGRAM:BINARY
:Disp "NUMBER TO"
:Disp "CONVERT:"
:Input A
:0→N
:0→B
:While 2^(N+1)≤A
:N+1→N
:End
:While N≥0
:iPart(A/2^N)→C
:10^(N)*C+B→B
:If C=1
:Then
:A-2^N→A
:End
:N-1→N
:End
:Disp B

Alternate using a string to display larger numbers. 

PROGRAM:BINARY
:Input X
:" "→Str1
:Repeat X=0
   :X/2→X
   :sub("01",2fPart(X)+1,1)+Str1→Str1
   :iPart(X)→X
:End
:Str1

Using the baseInput() "real(25," function from Omnicalc 

PROGRAM:BINARY
:Disp "NUMBER TO"
:Disp "CONVERT"
:Input "Str1"
:Disp real(25,Str1,10,2)

[edit] UNIX Shell

# Define a function to output binary digits
tobinary() {
  # We use the bench calculator for our conversion
  echo "obase=2;$1"|bc
}
 
# Call the function with each of our values
tobinary 5
tobinary 50

[edit] Vedit macro language

This implementation reads the numeric values from user input and writes the converted binary values in the edit buffer. 

repeat (ALL) {
    #10 = Get_Num("Give a numeric value, -1 to end: ", STATLINE)
    if (#10 < 0) { break }
    Call("BINARY")
    Update()
}
return
 
:BINARY:
do {
    Num_Ins(#10 & 1, LEFT+NOCR)
    #10 = #10 >> 1
    Char(-1)
} while (#10 > 0)
EOL
Ins_Newline
Return

Example output when values 0, 1, 5, 50 and 9000 were entered: 

0
1
101
110010
10001100101000

[edit] Visual Basic .NET

Sub Main()
    Console.WriteLine("5: " & Convert.ToString(5, 2))
    Console.WriteLine("50: " & Convert.ToString(50, 2))
    Console.WriteLine("9000: " & Convert.ToString(9000, 2))
End Sub

Output: 

5: 101
50: 110010
9000: 10001100101000

[edit] Whitespace

This program prints binary numbers until the internal representation of the current integer overflows to -1; it will never do so on some interpreters. It is almost an exact duplicate of Count in octal#Whitespace. 

    
 
    
  		
 	  	
 
 	 	
   	 	 
	
     	
	   
 
  
 
   	
 
    	 
	 		 
	   	 
	 	     
 	  	
	  	
		 	
 
 
 
   	 
 
 
		 		
	
 	
 
 	 
 
   		
 
 
 

It was generated from the following pseudo-Assembly. 

push 0
; Increment indefinitely.
0:
    push -1 ; Sentinel value so the printer knows when to stop.
    copy 1
    call 1
    push 10
    ochr
    push 1
    add
    jump 0
; Get the binary digits on the stack in reverse order.
1:
    dup
    push 2
    mod
    swap
    push 2
    div
    push 0
    copy 1
    sub
    jn 1
    pop
; Print them.
2:
    dup
    jn 3 ; Stop at the sentinel.
    onum
    jump 2
3:
    pop
    ret

[edit] X86 Assembly

Translation of XPL0. Assemble with tasm, tlink /t 

        .model  tiny
        .code
        .486
        org     100h
start:  mov     ax, 5
        call    binout
        call    crlf
        mov     ax, 50
        call    binout
        call    crlf
        mov     ax, 9000
        call    binout
 
crlf:   mov     al, 0Dh         ;new line
        int     29h
        mov     al, 0Ah
        int     29h
        ret
 
binout: push    ax
        shr     ax, 1
        je      bo10
         call   binout
bo10:   pop     ax
        and     al, 01h
        or      al, '0'
        int     29h             ;display character
        ret
        end     start

Output: 

101
110010
10001100101000

[edit] XPL0

include c:\cxpl\codes;          \intrinsic code declarations
 
proc BinOut(N);                 \Output N in binary
int N;
int R;
[R:= N&1;
N:= N>>1;
if N then BinOut(N);
ChOut(0, R+^0);
];
 
int I;
[I:= 0;
repeat  BinOut(I); CrLf(0);
        I:= I+1;
until   KeyHit or I=0;
]

Example output: 

0
1
10
11
100
101
110
111
1000
...
100000010011110
100000010011111
100000010100000
100000010100001

[edit] ZX Spectrum Basic

10 LET n=5: GO SUB 1000: PRINT s$
20 LET n=50: GO SUB 1000: PRINT s$
30 LET n=9000: GO SUB 1000: PRINT s$
999 STOP
1000 REM convert to binary
1010 LET t=n: REM temporary variable
1020 LET s$="": REM this will contain our binary digits
1030 LET sf=0: REM output has not started yet
1040 FOR l=126 TO 0 STEP -1
1050 LET d$="0": REM assume next digit is zero
1060 IF t>=(2^l) THEN LET d$="1": LET t=t-(2^l): LET sf=1
1070 IF (sf <> 0) THEN LET s$=s$+d$
1080 NEXT l
1090 RETURN
Retrieved from "http://rosettacode.org/mw/index.php?title=Binary_digits&oldid=158797"
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