Here's one attempt, assuming I got the requirements right:
int topbit(unsigned int x)
{
for (int i = CHAR_BIT * sizeof x - 1; i >= 0; --i)
{
if (x & (1u << i))
return i;
}
return -1;
}
unsigned int alignedxor(unsigned int a, unsigned int b)
{
const int topa = topbit(a);
const int topb = topbit(b);
if (topa < 0)
return b;
if (topb < 0)
return a;
if (topa > topb)
return a ^ (b << (topa - topb));
return (a << (topb - topa)) ^ b;
}
int main(void) {
printf("%x\n", alignedxor(0xce, 6));
printf("%x\n", alignedxor(6, 0xce));
return 0;
}
This prints e
, twice, which seems correct but that's all the testing I did.
And yes, you can get the index of the topmost 1-bit more efficiently, but who cares? Also used my rich imagination to deal with corner cases (such as one number being 0).