Question

What is the best solution for getting the base 2 logarithm of a number that I know is a power of two (2^k). (Of course I know only the value 2^k not k itself.)

One way I thought of doing is by subtracting 1 and then doing a bitcount:

lg2(n) = bitcount( n - 1 ) = k, iff k is an integer
0b10000 - 1 = 0b01111, bitcount(0b01111) = 4

But is there a faster way of doing it (without caching)? Also something that doesn’t involve bitcount about as fast would be nice to know?

One of the applications this is:

suppose you have bitmask
0b0110111000

and value
0b0101010101

and you are interested of
(value & bitmask) >> number of zeros in front of bitmask
(0b0101010101 & 0b0110111000) >> 3 = 0b100010

this can be done with

using bitcount
value & bitmask >> bitcount((bitmask - 1) xor bitmask) - 1

or using lg2
value & bitmask >> lg2(((bitmask - 1) xor bitmask) + 1 ) - 2

For it to be faster than bitcount without caching it should be faster than O(lg(k)) where k is the count of storage bits.

Answer 1

Many architectures have a “find first one” instruction (bsr, clz, bfffo, cntlzw, etc.) which will be much faster than bit-counting approaches.