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Programming-Idioms

  • Python
  • Pascal
  • Dart

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

var c = i.toRadixString(2).replaceAll('0','').length;
c = i.bit_count()

For Python 3.10+
c = bin(i).count("1")
c := PopCnt(i);
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;

Not optimized at all.
#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);

add even and odd bits
then add even and odd pairs of bits
then add even and odd quadruples of bits
then add even and odd octets of bits
then add whatever groups of 16 bits are called
done

with gcc you can also use the function _builtin_popcount
New implementation...
< > ⌨
deleplace 
var c = i.toRadixString(2).replaceAll('0','').length;
c = i.bit_count()
c = bin(i).count("1")
c := PopCnt(i);
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;
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 = 0;
for (; i; i &= i - 1, ++c);
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;
c = popcnt(i)
c := bits.OnesCount(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 = Data.Bits.popCount i

const c = i.toString(2).replace(/[^1]/g, '').length
int c = Integer.bitCount(i);
int c = new BigInteger(valueOf(i), 2).bitCount();
(setf c (logcount i))
$i = base_convert($i,10,2);
$x = str_replace(0,'',$i);
$c = strlen($x);
# Not best, but simple:
$c=0;
while ($i) {
   $c += $i&1;
   $i /= 2;
}
$c = unpack '%b*', pack 'i', $i;
c = i.digits(2).count(1)
let c = i.count_ones();