> Not necessarily

Fair. I should have said "when it doesn't *optimize*, there's now an
additional load that happens" instead of "vectorize". I was more thinking
about the -O0 cases, which I acknowledge talking about a few extra
operations when optimizations are turned off isn't the most important use
case to think about.

On Thu, Oct 31, 2024 at 3:26 PM Jonathan Wakely <cxx_at_[hidden]> wrote:

>
>
> On Thu, 31 Oct 2024 at 20:05, Matthew Kolbe via Std-Proposals <
> std-proposals_at_[hidden]> wrote:
>
>> > You can tell the compiler that this is safe either by loading both
>> values before the if
>>
>> As I mentioned in my other email, this is true (and I believe the best
>> work around). My two objections are that it's now two more lines of code.
>> The other is that taking both paths is no longer an available optimization,
>> but instead part of the code. So, even when it doesn't vectorize, there's
>> now an additional load that happens for all compilations.
>>
>
>
> Not necessarily: https://godbolt.org/z/4h8eoe6Ea
>
> The optimized code is identical whether you do two loads and only use one
> of them, or only do one load. In the case where you do two loads, one is
> dead code that can be removed. But it still tells the compiler that
> performing that load is not undefined, so the indices i and i+1 must be
> within the array bounds.
>
>

-- 
Matthew P. Kolbe
(312) 218-6595
matthew.kolbe_at_[hidden]