# first 1 in a bitmask using log2

I am trying to get the last 1 in a bitmask. More mathematically speaking, I have a number k, that can be written in its binary form as a sequence of 1 and 0. I want the "weight" or "index" of the last 1 in that number (from the right).

As an example, for the number 0b 10 0100 (36 in decimal), I want to get 3 (because the rightmost 1 is on 3rd position from the right)

I can find the first 1 easily (in my case 6), as the size of the number, or its base-2 log (rounded down +1), but what about the last one?

For information, I will use this in a dirty trick in MS Excel, described in http://stackoverflow.com/questions/5065432/excel-find-last-value-in-an-array/11246728#11246728

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A straightforward but not particularly nice solution is to divide repeatedly by $2$ until you get a non-integer; the number of divisions is the desired result. –  Brian M. Scott Jun 28 '12 at 14:53
If you have a reasonably conventional set of bitwise operators available, you can isolate the last bit using x AND NOT (x-1). But you might need to use logarithms to find its position. –  Henning Makholm Jun 28 '12 at 15:10

If $x$ is your original number, (~x + 1) & x will have only one bit set, the same as the least significant set bit of $x$. For example, if $x$ is 10101000, then (~x + 1) & x = 00001000. Perhaps that helps? You can then use a lookup table, or one of the techniques in Which bit is set to figure out which bit is still set.
Of course, in 2's complement arithmetic (~x + 1) is simply -x. –  Henning Makholm Jun 28 '12 at 15:17