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： > > On Wed, Jul 13, 2022 at 12:53 PM ChiaEn Wu <peterwu.pub@xxxxxxxxx> wrote: > > > Andy Shevchenko <andy.shevchenko@xxxxxxxxx> 於 2022年7月5日 週二 清晨5:14寫道： > > > > On Mon, Jul 4, 2022 at 7:43 AM ChiaEn Wu <peterwu.pub@xxxxxxxxx> wrote: Please, once again, remove unneeded context when replying! ^^^^^^^ ... > > > > > + prop_val = (ilog2(roundup_pow_of_two(prop_val)) + 1) >> 1; > > > > > > > > Isn't something closer to get_order() or fls()? > > > > > > I will revise it to "(get_order(prop_va * PAGE_SIZE) + 1) / 2" and > > > this change is meet your expectations?? > > > > Nope. Try again. What about fls()? > > 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? > // 1. use fls() and ffs() combination > prop_val = ffs(prop_val) == fls(prop_val) ? fls(prop_val) >> 1 : > (fls(prop_val) + 1) >> 1; > > // 2. use one line for-loop, but without fls() > for (i = --prop_val, prop_val = 0; i >> 2 * prop_val != 0; prop_val++); > ------------------------------------------------------------- > Do these changes meet your expectations?? No, this is ugly. Yes, I understand that a bit arithmetics is hard... -- With Best Regards, Andy Shevchenko
