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

On 07. 03. 25, 12:38, Ingo Molnar wrote:
> 
> * 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.

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...

thanks,
-- 
js
suse labs