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

Question

If I have some integer n, and I want to know the position of the most significant bit (that is, if the least significant bit is on the right, I want to know the position of

Answer 1

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

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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