exercise2-6.md

Exercise 2-6

Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.

Process

This one took about a week to wrap my head around bitwise operations. It was quite confusing in the beginning, however as I had a look at it everyday, it started becoming more intuitive, until the solution became obvious

The idea is:

1. replace the first number x field of length n at position p with just zeros
2. mask out only the field of length n of the second number y and move it to position p
3. combine both numbers using the or operator |

So for instance

1. x = 255 which is 1111 1111 in binary
2. then we say from position p = 5 for length of n = 4 replace with 0's which is 1100 0011 in binary
3. then grab the second number y = 85 which is ‭01010101‬ in binary
4. mask out only the field of n and move over into position p which is 0001 0100 in binary
5. finally, combine both numbers 1100 0011 and 0001 0100 using |
6. final answer is 1101 0111

Final Code

#include <stdio.h>

unsigned int setbits(unsigned int x, unsigned int p, unsigned int n, unsigned int y);

int main(void) {
    unsigned int x = 128;
    unsigned int p = 3;
    unsigned int n = 2;
    unsigned int y = 3;
    
    printf("x: %d\np: %d\nn: %d\ny: %d\n\n", x, p, n, y);
    printf("Result: %d\n\n", setbits(x, p, n, y));
    
    x = 255; p = 5; n = 6; y = 85;
    printf("x: %d\np: %d\nn: %d\ny: %d\n\n", x, p, n, y);
    printf("Result: %d\n", setbits(x, p, n, y));
    
    return 0;
}

unsigned int setbits(unsigned int x, unsigned int p, unsigned int n, unsigned int y) {
    unsigned int old = x & ~(~(~0 << n) << (p + 1 - n));    /* replace field for replacement with 0's */
    unsigned int new = (y & ~(~0 << n)) << (p + 1 - n);     /* get only rightmost field and bitshift into position */
    printf("Old: %d\nNew: %d\n", old, new);
    return old | new; /* combine old and new */
}

Output

Back to Main