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

Thanks!

Here is something that works with both negative and positive numbers --

int number_of_bits(int input)
{
    int nAnder = 1;
    int nCount = 0;
    while( nAnder )
    {
         if( input & nAnder )
              nCount++; 
         nAnder = nAnder << 1; 
    } 
    return nCount; 
}

I feel like there might be a way to add the binary digits ie. for 7 => 0111, 0+1+1+1

numberOfOnes(7){
digitArray[] = split(toBinary(7));
for(int i = 0; i < sizeOf(digitArray); i++){
result += digitArray[i];
return result;
}
}

i guess this also just works fine...
#include<iostream>
#include<stdio.h>

int main()
{
int n;
printf("\n Enter the number : ");
scanf("%d",&n);
int x=0;
// main logic
while(n!=0)
{
if(n%2!=0)
x++;
n=n/2;
}

printf("\n no. of binary one's is : %d \n",x);

return 0;
}

int numberOf1s(int n)
{
   int count = 0;
   if(n==0)return  count;
   while(n&=n-1)count++;
   return ++count;
}

void main()
{
  printf("%d",numberOf1s(7));
}

Output: 3
Works even for negative numbers, tried it!

Use Look-up Table to improve the performance.

int countOnes(unsigned int num){
	int count = 0;
	while(num){
		num &= (num-1);
		count++;
	}
	return count;
}

One of the easy approaches of the above algorithm is to check when the & operation of the given number with 1 is 1 because at that case only 1 bit will be the number where this condition is satisfied and elsewhere it will be 0 so, at the former case increment the counter variable and at the end return the count

Implementation:

#include<bits/stdc++.h>
using namespace std;
int countbits(int n){
int count = 0;
if(n == 0)
return 0;
while(n){
if(n&1 == 1)
count++;
n = n>>1;
}
return count;

}