Write a func invert(x,p,n) that returns x with the n bits that begin at position p inverted (1 becomes 0 & vice-versa), leaving the others unchanged

Question

Is the following solution to the question correct?

My solution is:

#include <stdio.h>

unsigned invert(unsigned x, int p, int n)
{
    return (x >> (p + 1 - n)) ^ ~(~0 << n);
}

int main()
{ 
    printf("%d\n", invert(6, 4, 3));
    return 0;
}

which prints the output 6, the binary equivalent of 0110.

Answer 1

No, it's not correct.

x >> (p+1-n) will shift all the bits of the original number, so that the most significant ones can't remain unchanged.

I'd use something like the following

unsigned invert(unsigned x, int p, int n)
{
    int const n_bits = sizeof(x) * CHAR_BIT;
    assert(p >= 0  &&  p < n_bits);
    assert(n >= 0  &&  n + p <= n_bits);
    if ( n == 0 )
        return x;
    unsigned const mask = (~0u >> (n_bits - n)) << p;

    return x ^ mask;
}