Exercise 2-7. Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged.

bitplay.c

#include    <stdio.h>

unsigned int invert(unsigned int x, int p, int n);

int main()
{
    unsigned int x;
    unsigned int y;
    unsigned int z;

    x = 255;
    y = invert(x, 0, 1);
    z = invert(y, 0, 1);

    fprintf(stdout, "x is %lu, inverted is %lu, inverted again is %lu\n", x, y, z);

    x = 0xbc;
    y = invert(x, 0, 1);
    z = invert(y, 0, 1);

    fprintf(stdout, "x is %lu, inverted is %lu, inverted again is %lu\n", x, y, z);

    // add your own test cases - at least 4 more.
    x = 255;
    y = invert(x, 2, 2);
    z = invert(y, 2, 2);

    fprintf(stdout, "x is %lu, inverted is %lu, inverted again is %lu\n", x, y, z);

    x = 255;
    y = invert(x, 4, 3);
    z = invert(y, 4, 3);

    fprintf(stdout, "x is %lu, inverted is %lu, inverted again is %lu\n", x, y, z);

    x = 255;
    y = invert(x, 7, 8);
    z = invert(y, 7, 8);

    fprintf(stdout, "x is %lu, inverted is %lu, inverted again is %lu\n", x, y, z);

    x = 0;
    y = invert(x, 7, 8);
    z = invert(y, 7, 8);

    fprintf(stdout, "x is %lu, inverted is %lu, inverted again is %lu\n", x, y, z);
}

unsigned int invert(unsigned int x, int p, int n)
{
    // places zeros in the rightmost n bits and make mask with ones in the rightmost n bits
    unsigned int mask = ((~(~0 << n)));

    // n-bit field of x that begins at position p
    unsigned int shift_no = p + 1 - n;

    // shift mask so that field to be retained at correct position 
    unsigned int shifted_mask = (mask << shift_no);

    // toggle original value so that only necessary values can change
    return (x ^ shifted_mask);
}