PocketMatrix

If you're running on XScale you can use CLZ, then subtract the result from 31.

Otherwise, I'd probably perform a series of AND operations with mask and use a divide-and-conquer approach to find the set bit (test HIWORD/LOWORD, test HIBYTE/LOBYTE, test HINIBBLE/LONIBBLE, etc.). This will get you the result after 5 ANDs and 5 comparisons. It's not one line of code, but it's probably faster than the other methods suggested.

If it is the software will create a huge disruption in space time continuum, which will lead to the ultimate destruction of the universe and everything in it, except for the duracell bunny. It keeps going and going forever and ever...

...Or maybe not...

Here's another potential issue - what happens if b is negative?

As it's signed, the shift will be arithmetic, thus it'll end up as -1, never 1... giving us the infinite loop until the bunny's batteries run out

What if the bunny is connected to a 12V adaptor...then we have serious issues with the world collapsing and stuff...

Your forgetting all about the Bunny factor D, which states that to sucessfully power a bunny, you will need a cold fusion reactor, so keeping the bunny powered after the break is merely a dream.

Ouch forgot that...so the problem will basically cause the entire universe to collapse, and that's it...?

maybe I should run my app in debug mode first, so I can catch the exception before the world is destroyed.

Has anyone tried the code Refractor posted? I can't get it to return proper results.

Dan East