Assuming that x is a 32 bit integer:
n = ((x >> 1) & 0x55555555) + (x & 0x55555555)
n = ((n >> 2) & 0x33333333) + (n % 0x33333333)
n = ((n >> 4) & 0x0F0F0F0F) + (n & 0x0F0F0F0F)
n = ((n >> 8) & 0x00FF00FF) + (n & 0x00FF00FF)
n = ((n >>16) & 0x0000FFFF) + (n & 0x0000FFFF)
n now is the number of bits set in x.
Dave
On Jul 3, 11:27 am, jalaj jaiswal <jalaj.jaiswa...@gmail.com> wrote:
> is there any better way of finding number of 1's in binary of a number other
> then below:
>
> #include<stdio.h>
> #include<stdlib.h>
> int main(){
> int n;
> printf("enter numb\n");
> scanf("%d",&n);
> int i=1;
> int count=0;
> for(int j=0;j<31;j++){
> if(n&(i<<j)){
> count++;
> }
> }
> printf("%d",count);
> system("pause");
> return 0;
>
> }
>
> --
> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
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