are you saying that x finally contains the number of bits that are set to
1..??

On Tue, Jul 12, 2011 at 1:09 AM, Dave <[email protected]> wrote:

> @rShetty: Ask a question. What do you need to know?
>
> Dave
>
> On Jul 11, 1:26 pm, rShetty <[email protected]> wrote:
> > Some more Explanation of the working would be helpful
> >
> > Thank You ..
> >
> > On Jul 11, 11:11 pm, Dave <[email protected]> wrote:
> >
> >
> >
> > > Assuming that the integer is 32 bits, this is pretty good:
> >
> > > x = (x & 0x55555555) +  ((x >> 1) & 0x55555555);
> > > x = (x & 0x33333333) +  ((x >> 2) & 0x33333333);
> > > x = (x & 0x0F0F0F0F) +  ((x >> 4) & 0x0F0F0F0F);
> > > x = (x & 0x00FF00FF) +  ((x >> 8) & 0x00FF00FF);
> > > x = (x & 0x0000FFFF) +  ((x >> 16) & 0x0000FFFF);
> >
> > > Notice that the hex constants are respectively alternate bits,
> > > alternate pairs of bits, alternate groups of four bits, alternate
> > > bytes, and the low-order half of the int.
> >
> > > The first statement determines the number of one-bits in each pair of
> > > bits. The second statement adds adjacent pairs of bits to get the
> > > number of bits in each group of four bits. Then these are added to get
> > > the number of bits in each byte, short int, and finally in the whole
> > > int.
> >
> > > Dave
> >
> > > On Jul 11, 12:44 pm, rShetty <[email protected]> wrote:
> >
> > > > What is the most efficient algorithm to count the number of bits in
> an
> > > > unsigned integer ?
> > > > Explain your approach to the problem ?- Hide quoted text -
> >
> > - Show quoted text -
>
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