Computing the floor of log₂(x) using only bitwise operators in C

Question

For homework, using C, I'm supposed to make a program that finds the log base 2 of a number greater than 0 using only the operators ! ~ & ^ | + << >>. I know that I'm supposed to shift right a number of times, but I don't know how to keep track of the number of times without having any loops or ifs. I've been stuck on this question for days, so any help is appreciated.

int ilog2(int x) {
    x = x | (x >> 1);
    x = x | (x >> 2);
    x = x | (x >> 4);
    x = x | (x >> 8);
    x = x | (x >> 16);
}

This is what I have so far. I pass the most significant bit to the end.

Answer 1

This gets the floor of logbase2 of a number.

int ilog2(int x) {

    int i, j, k, l, m;
    x = x | (x >> 1);
    x = x | (x >> 2);
    x = x | (x >> 4);
    x = x | (x >> 8);
    x = x | (x >> 16);

    // i = 0x55555555 
    i = 0x55 | (0x55 << 8); 
    i = i | (i << 16);

    // j = 0x33333333 
    j = 0x33 | (0x33 << 8);
    j = j | (j << 16);

    // k = 0x0f0f0f0f 
    k = 0x0f | (0x0f << 8);
    k = k | (k << 16);

    // l = 0x00ff00ff 
    l = 0xff | (0xff << 16);

    // m = 0x0000ffff 
    m = 0xff | (0xff << 8);

    x = (x & i) + ((x >> 1) & i);
    x = (x & j) + ((x >> 2) & j);
    x = (x & k) + ((x >> 4) & k);
    x = (x & l) + ((x >> 8) & l);
    x = (x & m) + ((x >> 16) & m);
    x = x + ~0;
    return x; 
}