CareerCup

int num /*Given 16 bit integer*/
int count = 0; /*gives total no. of 1's in num*/
for(i=0;i<16;i++)
{
if((num>>i) && 1)
{
count++;
}
}

Most optimal solution would be -

int num /* given 16 bit integer */
int count = 0; /* total no. of 1's in num */
while (num & (num-1)) {
count++;
num = num & (num-1);
}
print(++count);

n is any number.

void numberofones (int n)
{
int i=0;

while (n)
{
n = n&(n-1);
i++;
};
printf (" Number of ones : %d\n",i);

}

#include<stdio.h>
int main()
{
short a = 65535;
int i=0,count=0;

for(i=0;i<16;i++)
{
if(a & (1<<i))
count ++;
}
printf("\nCount:%d",count);
return 0;
}

The number of 1s in the given number can be found by right shifting the binary number of the given integer and counting the number of 1s if it satisfies the condition that (n&1 == 1) else it is not a set bit.

Implementation:

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

}