On Mon, Jun 6, 2011 at 11:34 AM, Roland Scheidegger <srol...@vmware.com> wrote: > Am 05.06.2011 03:55, schrieb Benjamin Bellec: >> Le 05/06/2011 03:05, Matt Turner a écrit : >>> On Sat, Jun 4, 2011 at 7:14 PM, Benjamin Bellec <b.bel...@gmail.com> wrote: >>>> Le 03/06/2011 06:09, Matt Turner a écrit : >>>>> Also, if you want to check if the value is already a power-of-two, >>>>> instead of a case statement for every POT (including 0), just do the >>>>> standard is-power-of-two check: >>>>> >>>>> (x & (x - 1)) == 0 >>>> >>>> My own tests (on a Core2) shows that it's less efficient to do that, at >>>> least with -O2 optimization enabled. With -O0, it's equal. >>> >>> For what input set? Powers of two? >> Both, my test case loops with 29 POT and 6 NPOT. >> I'm doing this because the OpenGL games that I have tested call the >> function more often with "good" values. >> >>> >>> Doesn't really matter, since the function isn't a hot path or >>> anything, but I'd suppose that the Linux kernel has its >>> is_power_of_2() function for a reason--that it's pretty ugly to have >>> lots of case statements like powers of two. >>> >>> Matt >> Ok, so here is a v3 patch which replace the switch statement. > > I like this one better too. > Do we actually need the x == 0 special case or would it be ok if we just > remove that (which will return 0)? > > Roland
It depends on the behavior you want. (x & (x - 1)) recognizes 0 as a power of two, so it would just return zero. I don't really know under what circumstances 0 would be input to this function. FYI, the function in the kernel has a special case that causes it to not detect 0 as a power of two. Matt _______________________________________________ mesa-dev mailing list mesa-dev@lists.freedesktop.org http://lists.freedesktop.org/mailman/listinfo/mesa-dev
