Question

Possible Duplicate:
Compute fast log base 2 ceiling

What is the fastest possible way to find out how many binary digits a particular integer has when it is converted from decimal to binary in C/C++?

Ex. 47₍₁₀₎ = 101111₍₂₎

So 47 has 6 digits represented in binary.

Answer 1

The fastest solution that’s presented at my favorite collection of bit twiddling hacks is Find the log base 2 of an N-bit integer in O(lg(N)) operations with multiply and lookup. It requires 13 instructions to find the highest set bit in a number.

uint32_t v; // find the log base 2 of 32-bit v
int r;      // result goes here

static const int MultiplyDeBruijnBitPosition[32] = 
{
  0, 9, 1, 10, 13, 21, 2, 29, 11, 14, 16, 18, 22, 25, 3, 30,
  8, 12, 20, 28, 15, 17, 24, 7, 19, 27, 23, 6, 26, 5, 4, 31
};

v |= v >> 1; // first round down to one less than a power of 2 
v |= v >> 2;
v |= v >> 4;
v |= v >> 8;
v |= v >> 16;

r = MultiplyDeBruijnBitPosition[(uint32_t)(v * 0x07C4ACDDU) >> 27];