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

H. Peter Anvin
2025-03-13T00:10:48Z

On March 12, 2025 4:56:31 PM PDT, Jacob Keller <jacob.e.keller@intel.com> wrote:
>
>
>On 3/7/2025 11:36 AM, David Laight wrote:
>> On Fri, 7 Mar 2025 12:42:41 +0100
>> Jiri Slaby <jirislaby@kernel.org> wrote:
>>
>>> On 07. 03. 25, 12:38, Ingo Molnar wrote:
>>>>
>>>> * Jiri Slaby <jirislaby@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@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.
>>>>
>>>> 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 me as well, I've made the same comment before.
>> When reading code I don't want to have to look up a function definition.
>> There is even scope for having parity_odd() and parity_even().
>> And, with the version that shifts a constant right you want to invert
>> the constant!
>>
>> David
>
>This is really a question of whether you expect odd or even parity as
>the "true" value. I think that would depend on context, and we may not
>reach a good consensus.
>
>I do agree that my brain would jump to "true is even, false is odd".
>However, I also agree returning the value as 0 for even and 1 for odd
>kind of made sense before, and updating this to be a bool and then
>requiring to switch all the callers is a bit obnoxious...

Odd = 1 = true is the only same definition. It is a bitwise XOR, or sum mod 1.