Pre-requisite: Input number n

Algorithm:

1)  Set count=1

2)  Do bit wise AND with n and 1. 
    n & 1

    let n be a7a6a5a4a3a2a1a0
    1->00000001
    So doing bitwise AND (refer to published article on bitwise operators) 
    will result in all bits 0 except the LSB which will be a0. 
    If a0 is 0 then the all bits are not set
    Thus,
    IF
        n& 1 == 1
        count++;
    END IF
    Right shift n by 1
    n=n>>1

3)  IF n==0
        Print count
    ELSE
        REPEAT step 1, 2

Example with Explanation:

Checking for 7
7->00000111

Initially, count=0

So first iteration:
n=7 //00000111
n & 1 =1
so , count=1
n=n>>1 (n=3) //00000011

So second iteration:
n=3 //00000011
n & 1 =1
count=2
n=n>>1 (n=1) //00000001

So third iteration:
n=1 //00000001
n & 1 =1
count=3
n=n>>1 (n=0) //00000000
So, Print count, 3

C implementation

#include <stdio.h>

int main()
{
	unsigned int n;
	printf("enter the integer\n");
	scanf("%d",&n);

	int count=0;

	while(n!=0){
		if(n & 1 == 1){ //if current bit 1
			count++;//increase count
		}
		n=n>>1;//right shift
	}

	printf("no of bits those are 1 ");
	printf("in its binary representation: %d\n",count);

	return 0;
}

Output

First run:
enter the integer
7
no of bits those are 1 in its binary representation: 3

Second run:
enter the integer
12
no of bits those are 1 in its binary representation: 2