ansi_c_pg47_ch02_ex06.c

/*
 * 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.
 */


#include <stdio.h>



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

	/*
	 * The mask will have the bits to be changed set and others unset. It would
	 * be easier if we are to generate this mask once and use it later on. Let
	 * us make a simple convention first. The left bits are all the bits left
	 * to the set bits. The set bits are the bits to be set by this function,
	 * and the right bits are on the right of the set bits.
	 *
	 * The mask should eventually look like
	 * 	00000000000000000000000111111111111111111110000000000000000000000000
	 *  |-------left bits-----||---set bits (n)---||--right bits (p+1-n)---|
	 *
	 * In order to get this mask we have to XOR the below components.
	 *
	 * The first component (~0 << (p+1-n)), should give us the below
	 * 	11111111111111111111111111111111111111111110000000000000000000000000
	 *  |-------left bits-----||---set bits (n)---||--right bits (p+1-n)---|
	 *
	 * The second component (~0<<(p+1), should give us the below
	 * 	11111111111111111111111000000000000000000000000000000000000000000000
	 *  |-------left bits-----||---set bits (n)---||--right bits (p+1-n)---|
	 */


	unsigned int mask = (~0 << (p+1-n)) ^ (~0<<(p+1));
	return (x&~mask)|((y<<(p+1-n))&mask);
}

int
main() {
	unsigned int x = 0xff;
	unsigned int y = 0x00;
	y = setbits(x,2,1,y);
	printf ("y = %X \n", y);
}