Create a mask that marks the most significant set bit, using only bitwise operators

Question

This was part of a larger programming assignment that was due for me last night. Couldn\'t figure out this problem, but I\'m curious as to how it could be solved.

Th

Answer 1

Updated to work for negative numbers (assuming that this should return 0x80000000 since these numbers have their top bit set )

int gbp(int n) {
 // return log(2) of n
 unsigned int m;
 m = n;
 m = m | m >> 1;
 m = m | m >> 2;
 m = m | m >> 4;
 m = m | m >> 8;
 m = m | m >> 16;
 m = m & ((~m >> 1)^0x80000000);
printf("m is now %d\n", m);
return m;
}

Explanation:

Starting with any bit pattern, when we shift right by 1 and take the OR, adjacent bits will become 1. Example

00010100
00001010
--------
00011110

You repeat this until you have all ones to the right of the leading digit, by successively shifting 2, 4, 8, 16 (if you have 32 bit numbers; for larger int you keep going).

Finally you need to "strip all the other ones" by inverting the number, right shifting by 1, and taking the AND:

00011111 AND 11110000 = 00010000

and there you have it.

For negative numbers, the final manipulation ensures that you don't kill the top bit if it exists. If you wanted something else to be done with negative numbers, let me know what it is.