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

Yury Norov
2025-03-13T17:14:58Z

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