Re: [PATCH 04/11] clk: divider: Introduce CLK_DIVIDER_EVEN_INTEGERS flag

[Théo Lebrun <theo.lebrun@xxxxxxxxxxx>]

Hello,

On Thu Apr 11, 2024 at 5:06 AM CEST, Stephen Boyd wrote:
> Quoting Théo Lebrun (2024-04-10 10:12:33)
> > index 4a537260f655..cb348e502e41 100644
> > --- a/include/linux/clk-provider.h
> > +++ b/include/linux/clk-provider.h
> > @@ -675,13 +675,15 @@ struct clk_div_table {
> >   * CLK_DIVIDER_BIG_ENDIAN - By default little endian register accesses are used
> >   *     for the divider register.  Setting this flag makes the register accesses
> >   *     big endian.
> > + * CLK_DIVIDER_EVEN_INTEGERS - clock divisor is 2, 4, 6, 8, 10, etc.
> > + *     Formula is 2 * (value read from hardware + 1).
> >   */
> >  struct clk_divider {
> >         struct clk_hw   hw;
> >         void __iomem    *reg;
> >         u8              shift;
> >         u8              width;
> > -       u8              flags;
> > +       u16             flags;
>
> This can stay u8

It is unclear to me why it can stay u8? __clk_hw_register_divider() puts
clk_divider_flags into flags field of struct clk_divider.
BIT(8) overflows u8.

>
> >         const struct clk_div_table      *table;
> >         spinlock_t      *lock;
> >  };
>
> We should add a kunit test.

Will look into how this works and try something for next revision. I
guess you are talking about adding clk_divider tests, not only tests
for this flag? I cannot find any existing kunit tests for clk_divider.

Thanks,

--
Théo Lebrun, Bootlin
Embedded Linux and Kernel engineering
https://bootlin.com