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

Jiri Slaby
2025-03-07T06:48:37Z

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@gmail.com>
> Signed-off-by: Yu-Chun Lin <eleanor15x@gmail.com>
> Signed-off-by: Kuan-Wei Chiu <visitorckw@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.

thanks,
--
js
suse labs