Compute fast log base 2 ceiling

Question

What is a fast way to compute the (long int) ceiling(log_2(i)), where the input and output are 64-bit integers? Solutions for signed or unsigned integers are ac

Answer 1

The code below is simpler and will work as long as the input x >= 1. input clog2(0) will get an undefined answer (which makes sense because log(0) is infinity...) You can add error checking for (x == 0) if you want:

unsigned int clog2 (unsigned int x)
{
    unsigned int result = 0;
    --x;
    while (x > 0) {
        ++result;
        x >>= 1;
    }

    return result;
}

By the way, the code for the floor of log2 is similar: (Again, assuming x >= 1)

unsigned int flog2 (unsigned int x)
{
    unsigned int result = 0;
    while (x > 1) {
        ++result;
        x >>= 1;
    }

    return result;
}