Hi Don,
instead of searching 1 at all places we can use this :
while(n!=0){
count++;
n = n & n-1;
}
On Fri, May 17, 2013 at 6:55 PM, Don <dondod...@gmail.com> wrote:
> Counting the set bits in one integer is not the problem which was
> asked.
> However, I think that something like this is both faster and more easy
> to understand than what you have written:
>
> int bitCount(unsigned int x)
> {
> int result = 0;
> while(x)
> {
> if (x & 1) ++result;
> x >>= 1;
> }
> return result;
> }
>
> On May 17, 8:32 am, bhargav <ybbkris...@gmail.com> wrote:
> > as bitwise operators are fast can count by following logic, works oly fr
> > +ve, just a tweak will make it to work with -ves also ..
> >
> > #include <stdio.h>
> > main() {
> > unsigned int x=12312,a;
> > a=x<<1;
> > //printf("%u",a);
> > int count=0;
> > while(x>0) {
> > a = x<<1;
> > //printf("%u \n",a);
> > if(a<x)
> > count++;
> > x=a;}
> >
> > printf("%d\n",count );
> > getch();
> >
> >
> >
> >
> >
> >
> >
> > }
>
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--
Mangal Dev
Ph No. 7411627654
Member Technical Staff
Oracle India Pvt Ltd
mangal....@oracle.com <mangal.d...@oracle.com>
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