Finding the exponent of n = 2**x using bitwise operations [logarithm in base 2 of n]

Question

Is there a straightforward way to extracting the exponent from a power of 2 using bitwise operations only?

EDIT: Although the question was originall

Answer 1

This is actually a comment to the performance test posted by mac. I post this as an answer to have proper code formatting and indenting

mac, could you try a unrolled implementation of the bitseach suggested by Mark Byers? Maybe it's just the array access that slows it down. In theory this approach should be faster than the others.

It would look something like this, although I'm not sure whether the formatting is right for python but I guess you can see what it is supposed to do.

def bitwise(v):
    r = 0;
    if( v > 0xffff ) : v >>= 16; r = 16;
    if( v > 0x00ff ) : v >>=  8; r += 8;
    if( v > 0x000f ) : v >>=  4; r += 4;
    if( v > 0x0003 ) : v >>=  2; r += 2;
    return r + ( v >> 1 );

If python shares java's lack of unsingned integers it would need to be something like this:

def bitwise(v):
    r = 0;
    if( v & 0xffff0000 ) : v >>>= 16; r = 16;
    if( v > 0x00ff ) : v >>=  8; r += 8;
    if( v > 0x000f ) : v >>=  4; r += 4;
    if( v > 0x0003 ) : v >>=  2; r += 2;
    return r + ( v >> 1 );