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

[Ingo Molnar]

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

> On 07. 03. 25, 12:38, Ingo Molnar wrote:
> > 
> > * 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.
> 
> Yes, that used to make sense for me. For bool/true/false, it no longer does.
> But as I wrote, it might be only me...

No strong opinion on this from me either, I'd guess existing practice 
with other parity functions should probably control. (If a coherent 
praxis exists.).

Thanks,

	Ingo