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

H. Peter Anvin
2025-03-13T17:15:03Z

On March 13, 2025 9:24:38 AM PDT, Yury Norov <yury.norov@gmail.com> wrote:
>On Wed, Mar 12, 2025 at 05:09:16PM -0700, H. Peter Anvin wrote:
>> On March 12, 2025 4:56:31 PM PDT, Jacob Keller <jacob.e.keller@intel.com> wrote:
>
>[...]
>
>> >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.
>
>The x86 implementation will be "popcnt(val) & 1", right? So if we
>choose to go with odd == false, we'll have to add an extra negation.
>So because it's a purely conventional thing, let's just pick a simpler
>one?
>
>Compiler's builtin parity() returns 1 for odd.
>
>Thanks,
>Yury

The x86 implementation, no, but there will be plenty of others having that exact definition.