Tuesday, August 26, 2014

Set or Unset a particular bit in an integer - A short program

Many times we need a method to set and unset a bit without having condition construct. It is easy to execute with 'if' construct. Just for fun, is it possible without having a conditional construct? Here is program I wrote

#include <stdio.h>
#include <stdint.h>
#include <stdbool.h>

set_unset_bit (uint32_t src, uint32_t pos, bool set)
    return ((src & ~((uint32_t)1 << (pos-1))) | (!!set << (pos-1)));

int main (int argc, char **argv)
    uint32_t test1 = 0;
    uint32_t test2 = 0xFFFFFFFF;

    printf("%x\n", set_unset_bit(test1, 10, 1));
    printf("%x\n", set_unset_bit(test2, 11, 0));

This works for uint32_t types and as expected.  What was the approach? Basically you need to set the bit of interest to zero and OR with whatever option user requested for (set or unset i.e. 1 or 0). The first part "(src & ~((uint32_t)1 << (pos-1)))" forcibly sets bit of requested position 'pos' to zero while the next construct "| (!!set << (pos-1))" sets or unsets based on the requirement. Hope you enjoy the small code snippet.

Why do you think '!!set' required? Read my previous article here


nandakumar@heramba ~/setbit $ ./a.out

Final note: This works only for uint32_t however can be expanded to other types too! For uint64_t, make sure to cast '1' to uint64_t!