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

From: Jacob Keller <jacob.e.keller@intel.com>
To: "H. Peter Anvin" <hpa@zytor.com>, Yury Norov <yury.norov@gmail.com>
Cc: David Laight <david.laight.linux@gmail.com>,
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Subject: Re: [PATCH v3 01/16] bitops: Change parity8() return type to bool
Date: Thu, 13 Mar 2025 14:09:24 -0700	[thread overview]
Message-ID: <b2b632cc-ca69-497f-9cf9-782bd02cac79@intel.com> (raw)
In-Reply-To: <795281B1-9B8A-477F-8012-DECD14CB53E5@zytor.com>

On 3/13/2025 9:36 AM, H. Peter Anvin wrote:
> 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.

Makes sense to stick with that existing convention then. Enough to
convince me.

Thanks,
Jake

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