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_form("/x2/\n", 9000);
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 <bitset>
#include <iostream>
#include <limits>
#include <string>
 
void print_bin(unsigned int n) {
std::string str = "0";
 
if (n > 0) {
str = std::bitset<std::numeric_limits<unsigned int>::digits>(n).to_string();
str = str.substr(str.find('1')); // remove leading zeros
}
 
std::cout << str << '\n';
}
 
int main() {
print_bin(0);
print_bin(5);
print_bin(50);
print_bin(9000);
}
 

Output:

0
101
110010
10001100101000

[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.
 

Free-form, using a reference modifier to index into binary-number.

IDENTIFICATION DIVISION.
PROGRAM-ID. binary-conversion.
 
DATA DIVISION.
WORKING-STORAGE SECTION.
01 binary-number pic X(21).
01 digit pic 9.
01 n pic 9(7).
01 nstr pic X(7).
01 ptr pic 99.
 
PROCEDURE DIVISION.
display "Number: " with no advancing.
accept nstr.
move nstr to n.
move zeroes to binary-number.
move length binary-number to ptr.
perform until n equal 0
divide n by 2 giving n remainder digit
move digit to binary-number(ptr:1)
subtract 1 from ptr
if ptr < 1
exit perform
end-if
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] Component Pascal

This example is incorrect. There shouldn't be leading zeros in the result. Please fix the code and remove this message.

BlackBox Component Builder

 
MODULE BinaryDigits;
IMPORT StdLog,Strings;
 
PROCEDURE Do*;
VAR
str : ARRAY 33 OF CHAR;
BEGIN
Strings.IntToStringForm(5,2,32,'0',FALSE,str);
StdLog.Int(5);StdLog.String(":> " + str);StdLog.Ln;
Strings.IntToStringForm(50,2,32,'0',FALSE,str);
StdLog.Int(50);StdLog.String(":> " + str);StdLog.Ln;
Strings.IntToStringForm(9000,2,32,'0',FALSE,str);
StdLog.Int(9000);StdLog.String(":> " + str);StdLog.Ln;
END Do;
END BinaryDigits.
 

Execute: ^Q BinaryDigits.Do
Output:

 5:> 00000000000000000000000000000101
 50:> 00000000000000000000000000110010
 9000:> 00000000000000000010001100101000

[edit] D

void main() {
import std.stdio;
 
foreach (immutable 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
function IntToBinStr(AInt : integer) : string;
begin
Result := '';
repeat
Result := Chr(Ord('0')+(AInt and 1))+Result;
AInt := AInt div 2;
until (AInt = 0);
end;
 
procedure Binary_Digits;
begin
writeln(' 5: '+IntToBinStr(5));
writeln(' 50: '+IntToBinStr(50));
writeln('9000: '+IntToBinStr(9000));
end;
   5: 101
  50: 110010
9000: 10001100101000

[edit] Elena

#define system.
#define extensions.
 
// --- Program ---
 
#symbol program =
[
console writeLine:(convertor toLiteral:5 &base:2).
console writeLine:(convertor toLiteral:50 &base:2).
console writeLine:(convertor toLiteral:9000 &base:2).
].

[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] FunL

for n <- [5, 50, 9000, 9000000000]
println( n, bin(n) )
Output:
5, 101
50, 110010
9000, 10001100101000
9000000000, 1000011000011100010001101000000000

[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] Idris

module Main
 
binaryDigit : Integer -> Char
binaryDigit n = if (mod n 2) == 1 then '1' else '0'
 
binaryString : Integer -> String
binaryString 0 = "0"
binaryString n = pack (loop n [])
where loop : Integer -> List Char -> List Char
loop 0 acc = acc
loop n acc = loop (div n 2) (binaryDigit n :: acc)
 
main : IO ()
main = do
putStrLn (binaryString 0)
putStrLn (binaryString 5)
putStrLn (binaryString 50)
putStrLn (binaryString 9000)
 

Output:

0
101
110010
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] jq

def binary_digits:
if . == 0 then "0"
else [recurse( if . == 0 then empty else ./2 | floor end ) % 2 | tostring]
| reverse
| .[1:] # remove the leading 0
| join("")
end ;
 
# The task:
(5, 50, 9000) | binary_digits
Output:
$ jq -n -r -f Binary_digits.jq
101
110010
10001100101000

[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] LFE

If one is simple printing the results and doesn't need to use them (e.g., assign them to any variables, etc.), this is very concise:

 
(: io format '"~.2B~n~.2B~n~.2B~n" (list 5 50 9000))
 

If, however, you do need to get the results from a function, you can use (: erlang integer_to_list ... ). Here's a simple example that does the same thing as the previous code:

 
(: lists foreach
(lambda (x)
(: io format
'"~s~n"
(list (: erlang integer_to_list x 2))))
(list 5 50 9000))
 

Both of these give the same 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

 
> convert( 50, 'binary' );
110010
> 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],
do (
[n, q]: divide(n, b),
v: cons(q, v),
if n=0 then return(v)))$
 
binary(n) := simplode(digits(n, 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] Nimrod

proc binDigits(x: BiggestInt, r: int): int =
## Calculates how many digits `x` has when each digit covers `r` bits.
result = 1
var y = x shr r
while y > 0:
y = y shr r
inc(result)
 
proc toBin*(x: BiggestInt, len: Natural = 0): string =
## converts `x` into its binary representation. The resulting string is
## always `len` characters long. By default the length is determined
## automatically. No leading ``0b`` prefix is generated.
var
mask: BiggestInt = 1
shift: BiggestInt = 0
len = if len == 0: binDigits(x, 1) else: len
result = newString(len)
for j in countdown(len-1, 0):
result[j] = chr(int((x and mask) shr shift) + ord('0'))
shift = shift + 1
mask = mask shl 1
 
for i in 0..15:
echo toBin(i)

Output:

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

[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 edit (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

@(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 simply.

[edit] simple version

Note: some REXX interpreters have a   D2B   [Decimal to Binary] BIF (built-in function).
Programming note:   this REXX version depends on numeric digits being large enough to handle leading zeroes in this manner (by added a zero (to the binary version) to force leading zero suppression).

/*REXX program demonstrates converting  decimal  ───►  binary.          */
numeric digits 1000
x.=
x.1 = 0
x.2 = 5
x.3 = 50
x.4 = 9000
do j=1 while x.j\=='' /*compute until a NULL is found.*/
y = x2b(d2x(x.j)) + 0 /*force removal of leading zeroes*/
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] elegant version

This version handles the case of zero as a special case more elegantly.
The following versions depend on the setting of numeric digits such that the number in decimal can be expressed as a whole number.

/*REXX program demonstrates converting  decimal  ───►  binary.          */
x.=
x.1 = 0
x.2 = 5
x.3 = 50
x.4 = 9000
do j=1 while x.j\=='' /*compute until a NULL is found.*/
y = strip( x2b( d2x( x.j )), 'L', 0)
if y=='' then y=0 /*handle special case of 0 (zero)*/
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] concise version

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

/*REXX program demonstrates converting  decimal  ───►  binary.          */
x.=
x.1=0
x.2=5
x.3=50
x.4=9000
do j=1 while x.j\=='' /*compute until a NULL is found.*/
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] conforming version

This REXX version conforms to the strict output requirements of this task (just show the binary output without any blanks).

/*REXX program demonstrates converting  decimal  ───►  binary.          */
numeric digits 200
x.=
x.1=0
x.2=5
x.3=50
x.4=9000
x.5=423785674235000123456789
x.6=1e138 /*one quinquaquadragintillion. */
 
do j=1 while x.j\=='' /*compute until a NULL is found.*/
y = strip( x2b( d2x( x.j )), 'L', 0)
if y=='' then y=0 /*handle special case of 0 (zero)*/
say y
end /*j*/
/*stick a fork in it, we're done.*/

output

0
101
110010
10001100101000
1011001101111010111011110101001101111000000000000110001100000100111110100010101
101010111111101001000101110110100000111011011011110111100110100100000100100001111101101110011101000101110110001101101000100100100110000111001010101011110010001111100011110100010101011011111111000110101110111100001011100111110000000010101100110101001010001001001011000000110000010010010100010010000001110100101000011111001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

[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] Rust

Works for Rust 0.11-pre.

fn main() {
for i in range(0, 16) {
println!("{:t}", i)
}
}

Outputs:

0
1
10
11
100
...

[edit] Scala

Scala has an implicit conversion from Int to RichInt which has a method toBinaryString.

scala> (5 toBinaryString)
res0: String = 101
 
scala> (50 toBinaryString)
res1: String = 110010
 
scala> (9000 toBinaryString)
res2: String = 10001100101000

[edit] Scheme

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

[edit] Seed7

This example uses the radix operator to write a number in binary.

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

Output:

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

[edit] Sidef

[5, 50, 9000].each { |n|
say n.to_bin.substr(2);
}
Output:
101
110010
10001100101000

[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] Swift

for num in [5, 50, 9000] {
println(String(num, radix: 2))
}
Output:
101
110010
10001100101000

[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] Vim Script

function Num2Bin(n)
let n = a:n
let s = ""
if n == 0
let s = "0"
else
while n
if n % 2 == 0
let s = "0" . s
else
let s = "1" . s
endif
let n = n / 2
endwhile
endif
return s
endfunction
 
echo Num2Bin(5)
echo Num2Bin(50)
echo Num2Bin(9000)
Output:
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] Wortel

Using JavaScripts buildin toString method on the Number object, the following function takes a number and returns a string with the binary representation:

\.toString 2
; the following function also casts the string to a number
^(@+ \.toString 2)

To output to the console:

@each ^(console.log \.toString 2) [5 50 900]
Outputs:
101
110010
1110000100

[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] zkl

(9000).toString(2)
T(5,50,9000).apply("toString",2) //--> L("101","110010","10001100101000")
"%.2B".fmt(9000)

[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
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