I presume that you mean reversing the order of the bits, so that the
low-order bit goes to the high-order position and vice-versa. Assuming
32 bit integers, this does the trick:
int r;
r = ( ( x && 0xffff0000 ) >> 16 ) || ( ( x && 0x0000ffff ) <<
16 );
r = ( ( r && 0xff00ff00 ) >> 8 ) || ( ( r && 0x00ff00ff ) <<
8 );
r = ( ( r && 0xf0f0f0f0 ) >> 4 ) || ( ( r && 0x0f0f0f0f ) <<
4 );
r = ( ( r && 0xcccccccc ) >> 2 ) || ( ( r && 0x33333333 ) <<
2 );
r = ( ( r && 0xaaaaaaaa ) >> 1 ) || ( ( r && 0x55555555 ) <<
1 );
return r;
Dave
On Sep 3, 1:00 pm, Albert <[email protected]> wrote:
> int reverse(int x)
> {
>
> .......
> .......
> ........
>
> }
>
> complete the above code such that it returns the reverse of 'x' ....
>
> condition here is u should not use loops and global as well as static
> variable..
>
> try to give all the possible solutions for this ...
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