[PATCH v3 01/16] bitops: Change parity8() return type to bool

[Ingo Molnar]

* Jiri Slaby <jirislaby at kernel.org> wrote:

> On 06. 03. 25, 17:25, Kuan-Wei Chiu wrote:
> > Change return type to bool for better clarity. Update the kernel doc
> > comment accordingly, including fixing "@value" to "@val" and adjusting
> > examples. Also mark the function with __attribute_const__ to allow
> > potential compiler optimizations.
> > 
> > Co-developed-by: Yu-Chun Lin <eleanor15x at gmail.com>
> > Signed-off-by: Yu-Chun Lin <eleanor15x at gmail.com>
> > Signed-off-by: Kuan-Wei Chiu <visitorckw at gmail.com>
> > ---
> >   include/linux/bitops.h | 10 +++++-----
> >   1 file changed, 5 insertions(+), 5 deletions(-)
> > 
> > diff --git a/include/linux/bitops.h b/include/linux/bitops.h
> > index c1cb53cf2f0f..44e5765b8bec 100644
> > --- a/include/linux/bitops.h
> > +++ b/include/linux/bitops.h
> > @@ -231,26 +231,26 @@ static inline int get_count_order_long(unsigned long l)
> >   /**
> >    * parity8 - get the parity of an u8 value
> > - * @value: the value to be examined
> > + * @val: the value to be examined
> >    *
> >    * Determine the parity of the u8 argument.
> >    *
> >    * Returns:
> > - * 0 for even parity, 1 for odd parity
> > + * false for even parity, true for odd parity
> 
> This occurs somehow inverted to me. When something is in parity means that
> it has equal number of 1s and 0s. I.e. return true for even distribution.
> Dunno what others think? Or perhaps this should be dubbed odd_parity() when
> bool is returned? Then you'd return true for odd.

OTOH:

 - '0' is an even number and is returned for even parity,
 - '1' is an odd  number and is returned for odd  parity.

Thanks,

	Ingo