Programming-Idioms

New implementation

Be concise.

Be useful.

All contributions dictatorially edited by webmasters to match personal tastes.

Please do not paste any copyright violating resource.

Please try to avoid dependencies to third-party libraries and frameworks.

Other implementations
#include <bitset>
std::bitset<sizeof(x)*8> y(x);
auto s = y.to_string();
String s = Convert.ToString(x,2);
String s = Convert.ToString(x,2).PadLeft(16, '0');
import std.conv;
auto s = to!string(x,2);
var s = x.toRadixString(2);
s = Integer.to_string(x, 2)
Integer.digits(x, 2) |> Enum.join("")
S = io_lib:format("~.2B~n", [X]).
write (unit=s,fmt='(B0)') x
write (unit=s,fmt='(B32)') x
import "fmt"
import "math/big"
s := fmt.Sprintf("%b", x)
import "strconv"
s := strconv.FormatInt(x, 2)
import Numeric (showIntAtBase)
import Data.Char (intToDigit)
s x = showIntAtBase 2 intToDigit x ""
import Data.List
foldl (\acc -> (++ acc) . show) "" . unfoldr (\n -> if x == 0 then Nothing else Just (x `mod` 2, x `div` 2))
let s = x.toString(2);
String s = Integer.toBinaryString(x);
local s = {}

while x > 0 do
    local tmp = math.fmod(x,2)
    s[#s+1] = tmp
    x=(x-tmp)/2
end

s=table.concat(s)
$s = sprintf("%b", $x);
StrUtils;
var
  _x: Integer;
  _s: String;
begin
  _s := IntToBin(_x,8*SizeOf(Integer));
end.
var Iter,n:integer;
[...]
  S := '';
  for Iter := 0 to n do
    s:= Char(Ord('0')+(x shr Iter) and 1) + S;   
$s = sprintf "%b", $x;
s = '{:b}'.format(x)
s = x.to_s(2)
let s = format!("{:b}", x);