chapter_2, p66.c

[E3798211]

This code does not work properly when x contains most significant bit set to 1, e.g. 0xFFFFFFFF.
Maybe this solution might avoid this problem:

0. Main idea is to shift right and add 0x1 to set bit
1. Save if (x == 0) to return 0 in this case: unsigned add_bit = (x == 0);
2. After setting all right bits to 1, shift right: x >>= 1;
3. Add bit: return (x + add_bit);

So the code will look like this:
int leftmost_one(unsigned x)
{
unsigned add_bit = (x != 0);
x |= (x >> 1);
x |= (x >> 2);
x |= (x >> 4);
x |= (x >> 8);
x |= (x >> 16);
x >>= 1;
return (x + add_bit); // Add 1 if x != 0, 0 otherwise
}

Thanks!