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Idiom #84 Count bits set in integer binary representation

Count number c of 1s in the integer i in base 2.

E.g. i=6 → c=2

#include <stdint.h>
uint32_t c = i;
c = (c & 0x55555555) + ((c & 0xAAAAAAAA) >> 1);
c = (c & 0x33333333) + ((c & 0xCCCCCCCC) >> 2);
c = (c & 0x0F0F0F0F) + ((c & 0xF0F0F0F0) >> 4);
c = (c & 0x00FF00FF) + ((c & 0xFF00FF00) >> 8);
c = (c & 0x0000FFFF) + ((c & 0xFFFF0000) >> 16);
(def c (count (re-seq #"1" (Integer/toBinaryString i))))
#include <cstdint>
uint32_t c = i;
c = (c & 0x55555555) + ((c & 0xAAAAAAAA) >> 1);
c = (c & 0x33333333) + ((c & 0xCCCCCCCC) >> 2);
c = (c & 0x0F0F0F0F) + ((c & 0xF0F0F0F0) >> 4);
c = (c & 0x00FF00FF) + ((c & 0xFF00FF00) >> 8);
c = (c & 0x0000FFFF) + ((c & 0xFFFF0000) >> 16);
public static int BitCount(int i)
{
    var c = 0;
    while (n != 0)
    {
        c++;
        n &= (n - 1); //walking through all the bits which are set to one
    }

    return c;
}
using System.Numerics;
var c = BitOperations.PopCount((uint)i);
import std.bitmanip;
auto c = bitSet(i).length;
var c = i.toRadixString(2).replaceAll('0','').length;
c = popcnt(i)
func PopCountUInt64(i uint64) (c int) {
	i -= (i >> 1) & 0x5555555555555555
	i = (i>>2)&0x3333333333333333 + i&0x3333333333333333
	i += i >> 4
	i &= 0x0f0f0f0f0f0f0f0f
	i *= 0x0101010101010101
	return int(i >> 56)
}

func PopCountUInt32(i uint32) (n int) {
	i -= (i >> 1) & 0x55555555
	i = (i>>2)&0x33333333 + i&0x33333333
	i += i >> 4
	i &= 0x0f0f0f0f
	i *= 0x01010101
	return int(i >> 24)
}
import "math/bits"
c := bits.OnesCount(i)
c = Data.Bits.popCount i

const c = i.toString(2).replace(/[^1]/g, '').length
int c = Integer.bitCount(i);
(setf c (logcount i))
$i = base_convert($i,10,2);
$x = str_replace(0,'',$i);
$c = strlen($x);
function BitCount(N: Int64): Integer;
var
  Q: QWord;
  i: Integer;
begin
  Result := 0;
  Q := QWord(N);
  for i := 0 to (8 * SizeOf(N) - 1) do
  begin
    if ((Q and 1) = 1) then Inc(Result);
    Q := Q shr 1;
  end;
end;
c := PopCnt(i);
$c = unpack '%b*', pack 'i', $i;
# Not best, but simple:
$c=0;
while ($i) {
   $c += $i&1;
   $i /= 2;
}
c = bin(i).count("1")
c = i.digits(2).count(1)
let c = i.count_ones();
New implementation...
< > ⌨
deleplace 
uint32_t c = i;
c = (c & 0x55555555) + ((c & 0xAAAAAAAA) >> 1);
c = (c & 0x33333333) + ((c & 0xCCCCCCCC) >> 2);
c = (c & 0x0F0F0F0F) + ((c & 0xF0F0F0F0) >> 4);
c = (c & 0x00FF00FF) + ((c & 0xFF00FF00) >> 8);
c = (c & 0x0000FFFF) + ((c & 0xFFFF0000) >> 16);
(def c (count (re-seq #"1" (Integer/toBinaryString i))))
uint32_t c = i;
c = (c & 0x55555555) + ((c & 0xAAAAAAAA) >> 1);
c = (c & 0x33333333) + ((c & 0xCCCCCCCC) >> 2);
c = (c & 0x0F0F0F0F) + ((c & 0xF0F0F0F0) >> 4);
c = (c & 0x00FF00FF) + ((c & 0xFF00FF00) >> 8);
c = (c & 0x0000FFFF) + ((c & 0xFFFF0000) >> 16);
public static int BitCount(int i)
{
    var c = 0;
    while (n != 0)
    {
        c++;
        n &= (n - 1); //walking through all the bits which are set to one
    }

    return c;
}
var c = BitOperations.PopCount((uint)i);
auto c = bitSet(i).length;
var c = i.toRadixString(2).replaceAll('0','').length;
c = popcnt(i)
func PopCountUInt64(i uint64) (c int) {
	i -= (i >> 1) & 0x5555555555555555
	i = (i>>2)&0x3333333333333333 + i&0x3333333333333333
	i += i >> 4
	i &= 0x0f0f0f0f0f0f0f0f
	i *= 0x0101010101010101
	return int(i >> 56)
}

func PopCountUInt32(i uint32) (n int) {
	i -= (i >> 1) & 0x55555555
	i = (i>>2)&0x33333333 + i&0x33333333
	i += i >> 4
	i &= 0x0f0f0f0f
	i *= 0x01010101
	return int(i >> 24)
}
c := bits.OnesCount(i)
c = Data.Bits.popCount i

const c = i.toString(2).replace(/[^1]/g, '').length
int c = Integer.bitCount(i);
(setf c (logcount i))
$i = base_convert($i,10,2);
$x = str_replace(0,'',$i);
$c = strlen($x);
function BitCount(N: Int64): Integer;
var
  Q: QWord;
  i: Integer;
begin
  Result := 0;
  Q := QWord(N);
  for i := 0 to (8 * SizeOf(N) - 1) do
  begin
    if ((Q and 1) = 1) then Inc(Result);
    Q := Q shr 1;
  end;
end;
c := PopCnt(i);
$c = unpack '%b*', pack 'i', $i;
# Not best, but simple:
$c=0;
while ($i) {
   $c += $i&1;
   $i /= 2;
}
c = bin(i).count("1")
c = i.digits(2).count(1)
let c = i.count_ones();