Replying to myself: I should proofread better _before_ I post. Sorry,
but the explanation should say
ni is bit i of n. nj is bit j of n. n & ~(ni | nj) removes the two
bits, and then they are shifted to the exchanged positions and or'd
in.
Dave
On Sep 13, 2:04 pm, Dave <dave_and_da...@juno.com> wrote:
> @Kumar: How about this:
>
> int exchange2bits(int n, int i, int j) // exchange bits i and j of n.
> {
> int ni, nj;
> ni = n & (1 << i);
> nj = n & (1 << j);
> return n & ~(ni | nj) | ((ni >> i) << j) | ((nj >> j) << i);
>
> }
>
> ni is bit i of n. nj is bit j of n. n & (ni | nj) removes the two
> bits, and then they are shifted to the exchanged positions and or'd
> in.
>
> Dave
>
> 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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