int temp = {[1<<(j-+1)]<<i-1};
Here temp is a number with all the bits set between positions i & j [both
inclusive]
temp = ~temp;
N = N & temp; // here we are clearing all the bits of N from
position i to j
temp = temp | M; // now we are taking the bit pattern from M into temp
in the given positions
N = N | temp; // now again we are setting the same pattern from
temp into N.
Note :- clearing bit means bit set to zero , while setting bit means bit is
1
--
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To view this discussion on the web visit
https://groups.google.com/d/msg/algogeeks/-/QTXreMoSy6gJ.
To post to this group, send email to [email protected].
To unsubscribe from this group, send email to
[email protected].
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
