suppose we have to swap bits b1 and b2 in a number n.Then
if(b1==b2) do not do any thing
else take a hexadecimal number whose all bits are zero except bits b1 and
b2.bit b1=1 and bit b2=1
now xor of original no with this no will give the desired result.
On Sun, Oct 30, 2011 at 10:32 AM, shiva@Algo <shiv.jays...@gmail.com> wrote:
> we need to swap only if both the bits are not same
>
> if((n^(1<<i)<n)!=(n^(1<<j)<n))
> n^=(1<<i)+(1<<j);
>
> On 10/29/11, praveen raj <praveen0...@gmail.com> wrote:
> > int func(int x)
> > {
> > int y=(1<<i)+(1<<j);
> > int z=x&y; // if after bitwise and ..we get power of 2 then
> ...
> > we have to flip the bits..
> > if((z&(z-1))==0)
> > return(x^y);
> > else
> > return x;
> > }
> >
> > With regards,
> >
> > Praveen Raj
> > DCE-IT
> > 9999735993
> > praveen0...@gmail.com
> >
> >>
> >>
> >
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