Re: [PATCH v3 01/16] bitops: Change parity8() return type to bool
Ingo Molnar <[email protected]>
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* Jiri Slaby <[email protected]> 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 <[email protected]> > > Signed-off-by: Yu-Chun Lin <[email protected]> > > Signed-off-by: Kuan-Wei Chiu <[email protected]> > > --- > > 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