C program to count bits set in an integer

Method1

This is the most basic way of doing it.

#include
int main()
{
unsinged int num=10;
int ctr=0;

for(;num!=0;num>>=1)
{
if(num&1)
{
ctr++;
}
}

printf("\n Number of bits set in [%d] = [%d]\n", num, ctr);
return(0);

}


Method2

This is a faster way of doing the same thing. Here the control goes into the while loop only as many times as the number of bits set to 1 in the integer!.

#include
int main()
{
int num=10;
int ctr=0;

while(num)
{
ctr++;
num = num & (num - 1); // This clears the least significant bit set.
}

printf("\n Number of bits set in [%d] = [%d]\n", num, ctr);
return(0);

}



Method3
This method is very popular because it uses a lookup table. This speeds up the computation. What it does is it keeps a table which hardcodes the number of bits set in each integer from 0 to 256.

For example

0 - 0 Bit(s) set.
1 - 1 Bit(s) set.
2 - 1 Bit(s) set.
3 - 2 Bit(s) set.
...


So here is the code...



const unsigned char LookupTable[] = { 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8};


unsigned int num;
unsigned int ctr; // Number of bits set.

ctr = LookupTable[num & 0xff] +
LookupTable[(num >> 8) & 0xff] +
LookupTable[(num >> 16) & 0xff] +
LoopupTable[num >> 24];