Some inaccuracies:
It's true that a binary search in a 64 bit unsigned integer range is going to take, at worst, 64 comparisons. ... This is O(log(n))

The initial proposal is a binary search on the bits which (as stated in the original post) takes at most 6 comparisons, not 64. It would be O(log(log(n)))

But a linear search through a 64-bit vector to find the most significant bit for an integer, is also O(log(n)); .... And a linear search will be very fast for such a small problem space.

Yes, linear on the number of bits, which is a loop of 64 comparisons, vs. the 6 proposed by OP.
So, slower.

Therefore, a solution that requires NO iteration at all could be ... though on paper this solution is O(1)

Hiding a loop inside a function doesn't make it not-a-loop. Since the conversation is about bits, you can't just assume it as a constant like when you're assuming a 64-bit hardware op. If this were an arbitrary precision number like Math::BigInt, 'log' is definitely not a constant operation.

In reply to Re^2: Most Significant Set Bit by NERDVANA
in thread Most Significant Set Bit by coldr3ality

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