On Thu, Jul 14, 2022 at 11:43 AM Andy Shevchenko <andy.shevchenko@xxxxxxxxx> wrote: > 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： ... > > > * 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; Okay, I nailed it down, but Daniel is right, it's simpler to have just conditionals. y = x >=2 ? __fls(x - 1) / 2 + 1 : 0; -- With Best Regards, Andy Shevchenko
