[PATCH v4 13/13] video: backlight: mt6370: Add Mediatek MT6370 support

[Andy Shevchenko]

On Thu, Jul 14, 2022 at 11:43 AM Andy Shevchenko
<andy.shevchenko at gmail.com> wrote:
> On Thu, Jul 14, 2022 at 11:27 AM Andy Shevchenko
> <andy.shevchenko at gmail.com> wrote:
> > On Thu, Jul 14, 2022 at 9:13 AM ChiaEn Wu <peterwu.pub at gmail.com> wrote:
> > > Andy Shevchenko <andy.shevchenko at gmail.com> 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