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#include <stdio.h>
/* The C Programming Language: 2nd Edition
*
* Exercise 2-07: 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.
*
* Answer: The trick is to start with ~0 to get all 1s, then left-shift n
* times. After that, flip again to get n number of 1s. Next, left shift the
* appropriate number of times needed to line up the mask of 1s to the proper
* location. Apply that mask to x with XOR and bam, job done!
*
* Here's a step by step illustration of invert(205, 5, 4), assuming 16-bit
* unsigned integers:
*
* 1111111111111111 ~0
* 1111111111110000 << 4
* 0000000000001111 ~
* 0000000000011110 << (5 - 4 == 1)
*
* Mask created. Now let's take our original number...
* 0000000011001101 205
* #### ^ ...and apply the mask!
* 0000000011010011 End Result (211)
*/
unsigned invert(unsigned x, unsigned p, unsigned n) {
return x ^ ((~(~0 << n)) << (p - n));
}
int main() {
printf("invert(205, 5, 4) returns %d\n", invert(205, 5, 4));
printf("11001101 flips 110[0110]1 to produce...\n11010011\n");
printf("invert(1876, 7, 3) returns %d\n", invert(1876, 7, 3));
printf("11101010100 flips 1110[101]0100 to produce...\n11100100100\n");
return 0;
}
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