Answer by Jonathan Mee for What is the fastest/most efficient way to find the highest set bit (msb) in an integer in C?

c99 has given us log2. This removes the need for all the special sauce log2 implementations you see on this page. You can use the standard's log2 implementation like this:

const auto n = 13UL;const auto Index = (unsigned long)log2(n);printf("MSB is: %u\n", Index); // Prints 3 (zero offset)

An n of 0UL needs to be guarded against as well, because:

-∞ is returned and FE_DIVBYZERO is raised

I have written an example with that check that arbitrarily sets Index to ULONG_MAX here: https://ideone.com/u26vsi

The visual-studio corollary to ephemient's gcc only answer is:

const auto n = 13UL;unsigned long Index;_BitScanReverse(&Index, n);printf("MSB is: %u\n", Index); // Prints 3 (zero offset)

The documentation for _BitScanReverse states that Index is:

Loaded with the bit position of the first set bit (1) found

In practice I've found that if n is 0UL that Index is set to 0UL, just as it would be for an n of 1UL. But the only thing guaranteed in the documentation in the case of an n of 0UL is that the return is:

0 if no set bits were found

Thus, similarly to the preferable log2 implementation above the return should be checked setting Index to a flagged value in this case. I've again written an example of using ULONG_MAX for this flag value here: http://rextester.com/GCU61409