I think we need a fix or a revert for this today, please. Bootstrap has been
broken
for a couple of days during the last week of stage 3, which is really
problematic.
Thanks,
Bill
On 1/12/22 6:57 AM, Richard Biener via Gcc-patches wrote:
> On Wed, 12 Jan 2022, Andre Vieira (lists) wrote:
>
>> On 12/01/2022 11:59, Richard Biener wrote:
>>> On Wed, 12 Jan 2022, Andre Vieira (lists) wrote:
>>>
>>>> On 12/01/2022 11:44, Richard Sandiford wrote:
>>>>> Another alternative would be to push autodetected_vector_mode when the
>>>>> length is 1 and keep 1 as the starting point.
>>>>>
>>>>> Richard
>>>> I'm guessing we would still want to skip epilogue vectorization if
>>>> !VECTOR_MODE_P (autodetected_vector_mode) in that case?
>>> Practically we currently only support fixed width word_mode there,
>>> but eventually one could end up with 64bit DImode for the main loop
>>> and 32bit V4QImode in the epilogue ... so not sure if it's worth
>>> special-casing. But I don't mind adding that skip.
>>>
>>> Richard.
>> I left out the skip, it shouldn't break anything as it would try that same
>> mode before anyway.
>> Just to clarify what I meant though was to skip if autodetected_vector_mode
>> wasn't a vector AND the target didn't define autovectorize_vector_modes, so
>> in
>> that scenario it wouldn't ever try V4QImode for the epilogue if the mainloop
>> was autodetected DImode, I think...
>> Either way, this is less code, less complicated and doesn't analyze more than
>> it did before the original patch, so I'm happy with that too.
>>
>> Is this what you had in mind?
> - mode_i = 1;
> + if (vector_modes.length () == 1)
> + {
> + /* If we only had VOIDmode then use AUTODETECTED_VECTOR_MODE to see
> if
> + an epilogue can be created with that mode. */
> + vector_modes[0] = autodetected_vector_mode;
> + mode_i = 0;
> + }
> + else
> + mode_i = 1;
> +
>
> I would have left out the condition and unconditionally do
>
> vector_modes[0] = autodetected_vector_mode;
> mode_i = 0;
>
> but OK if you think it makes sense to special case length == 1.
>
> Richard.
