On Thu, Jul 14, 2022 at 11:27 AM Andy Shevchenko <andy.shevchenko@xxxxxxxxx> wrote: > > On Thu, Jul 14, 2022 at 9:13 AM ChiaEn Wu <peterwu.pub@xxxxxxxxx> wrote: > > Andy Shevchenko <andy.shevchenko@xxxxxxxxx> 於 2022年7月13日 週三 晚上8:07寫道： ... > > I have tried two methods so far, as follows > > ------------------------------------------------------------- > > /* > > * prop_val = 1 --> 1 steps --> b'00 > > * prop_val = 2 ~ 4 --> 4 steps --> b'01 > > * prop_val = 5 ~ 16 --> 16 steps --> b'10 > > * prop_val = 17 ~ 64 --> 64 steps --> b'11 > > */ > > So, for 1 --> 0, for 2 --> 1, for 5 --> 2, and for 17 --> 3. > Now, consider x - 1: > 0 ( 0 ) --> 0 > 1 (2^0) --> 1 > 4 (2^2) --> 2 > 16 (2^4) --> 3 > 64 (2^6) --> ? (but let's consider that the range has been checked already) > > Since we take the lower limit, it means ffs(): > > y = (ffs(x - 1) + 1) / 2; > > Does it work for you? It wouldn't, because we need to use fls() against it actually. So, 0..1 (-1..0) --> 0 2..4 (1..3) --> 1 5..16 (4..15) --> 2 17..64 (16..63) --> 3 y = x ? ((fls(x - 1) + 1) / 2 : 0; -- With Best Regards, Andy Shevchenko
