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 77->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

Pre-requisite: Input numbernAlgorithm:Example with Explanation:C implementation

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