1. void draw(){
2.   println( frameCount + " " + binary(frameCount) + " " + lastOne(frameCount) );
3. }
4. int lastOne( int input ){
5.   return( input==0?-1:((input&1)!=0?0:1+lastOne(input>>1)) );
6. }

amnon.owed, I understand how that one works, it is probably a good solution for me.

  returnValue = valueToReturnIfConditionIsTrue;

} else {
  returnValue = valueToReturnIfConditionIsFalse;

}

Good suggestions so far, but the most efficient one has not been mentioned yet: use bit shifting =)

Want the most significant bit? swap the shifts:

Since you want to know about positions of bits, the fastest solution is to literally work with bits, with << and >> shifting all the bits in a value left or right by ... spots, so << 1 moves all the bits left by 1, << 8 by 8 positions, etc.

If we want to find the first high bit from the left or right, we simply move bits to the left or right in increasingly big steps, then moving them all back. bits that are "pushed off" are lost, so moving back puts zeroes in their place instead. As such, if we shift some bits left X positions, and then right X positions, and the result is not the same, we lost a 1 by moving X spots - the first 1 from the left or right is therefore at spot X.

You can also do this with a blind while, starting at p=1 and then increasing p by one, every step. However, I'd advise against this because this will cause an infinite loop when frameCount is 0 - since frameCount starts at 1, this sounds impossible, but in reality the frameCount value will overrun at some point (far, far in the future) so it could theoretically happen. So we use a for loop with a clear cutoff point (after 32 bits, because there's only 32 bits in an int) so that we don't bake an infinite loop into our program!

- Pomax