CareerCup

while(num)
{
    ctr++;
    num = num & (num-1);
}

for (int s=0,int n=0; n; n>>=1)
s+=n&1;

return s;

Or use kernighan's method for less iteration

int nos_of_ones(int n){
int count = 0;
while (n!=0){
if (n&1){
count++;
}
n>>1;
}
return (count);
}

int nos_of_ones(int n){
int count = 0;
while (n!=0){
if (n&1){
count++;
}
n>>1;
}
return (count);
}

int nos_of_ones(int n){
int count = 0;
while (n!=0){
if (n&1){
count++;
}
n>>1;
}
return (count);
}

int nCount=0/*number of binary 1's*/, i, input = 98/*some random number*/;
for(i=0; i<8*sizeof(input);i++)
{
if(input & (1<<i))
nCount++
}

count(int n)
{
int r,count=0;
while(n>0)
{
r=n%2;
if(r==1) count++;
n/=2;
}
return count;}

is it a must that we use bitwise operators? the above soln works too with O(n) complexity, n is the number of bits in binary rep.

int num_0 = 0;
while (n) {
          num_0 ++;
          n &= n-1;
}

Is it possible that we can optimize this solution by casting the number as a series of bytes? Using this we can iterate over each byte and can possibly optimize by skipping over bytes that are zero and run the same algo as mentioned before on each byte.