Yes, good catch. It should be the bitwise or.
Don
On Sep 13, 3:37 pm, Dave <dave_and_da...@juno.com> wrote:
> @Don. Very nice. I think you meant | where you wrote ||. This is so
> short that you could even do it as a #define:
>
> #define bitExchange(n,i,j) (((((n)>>(i))&1)==(((n)>>(j))&1))?n:
> (n)^((1<<(i))|(1<<(j))))
>
> Dave
>
> On Sep 13, 3:15 pm, Don <dondod...@gmail.com> wrote:
>
> > // If bits i and j are equal, result is just n
> > // Otherwise, toggle bits i and j
> > int bitExchange(int n, int i, int j)
> > {
> > return (((n>>i)&1) == ((n>>j)&1)) ? n : n ^ ((1<<i) || (1<<j));
>
> > }
>
> > On Sep 13, 1:50 pm, kumar raja <rajkumar.cs...@gmail.com> wrote:
>
> > > Suppose a number 'n' is given and two bits positions i,j present in
> > > binary
> > > representation of n .
>
> > > Then how to exchange the contents of the two bits i and j.
>
> > > E.g. n= 13
> > > its binary representation is 0000 1101 (just for now consider 8 bit
> > > number)
>
> > > i= 2,j=6
>
> > > o/p : 0100 1001 = 73
>
> > > please suggest some effective way to do this...
>
> > > --
> > > Regards
> > > Kumar Raja
> > > M.Tech(SIT)
> > > IIT Kharagpur,
> > > 10it60...@iitkgp.ac.in
> > > 7797137043.
> > > 09491690115.- Hide quoted text -
>
> > - Show quoted text -
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