Re: [PATCH 0/2] don't use mapcount() to check large folio sharing

Yin, Fengwei
2023-08-02T12:35:49Z

On 8/2/2023 6:27 PM, Ryan Roberts wrote:
> On 28/07/2023 17:13, Yin Fengwei wrote:
>> In madvise_cold_or_pageout_pte_range() and madvise_free_pte_range(),
>> folio_mapcount() is used to check whether the folio is shared. But it's
>> not correct as folio_mapcount() returns total mapcount of large folio.
>>
>> Use folio_estimated_sharers() here as the estimated number is enough.
>>
>> Yin Fengwei (2):
>> madvise: don't use mapcount() against large folio for sharing check
>> madvise: don't use mapcount() against large folio for sharing check
>>
>> mm/huge_memory.c | 2 +-
>> mm/madvise.c | 6 +++---
>> 2 files changed, 4 insertions(+), 4 deletions(-)
>>
>
> As a set of fixes, I agree this is definitely an improvement, so:
>
> Reviewed-By: Ryan Roberts
Thanks.

>
>
> But I have a couple of comments around further improvements;
>
> Once we have the scheme that David is working on to be able to provide precise
> exclusive vs shared info, we will probably want to move to that. Although that
> scheme will need access to the mm_struct of a process known to be mapping the
> folio. We have that info, but its not passed to folio_estimated_sharers() so we
> can't just reimplement folio_estimated_sharers() - we will need to rework these
> call sites again.
Yes. This could be extra work. Maybe should delay till David's work is done.

>
> Given the aspiration for most of the memory to be large folios going forwards,
> wouldn't it be better to avoid splitting the large folio where the large folio
> is mapped entirely within the range of the madvise operation? Sorry if this has
> already been discussed and decided against - I didn't follow the RFC too
> closely. Or perhaps you plan to do this as a follow up?
Yes. We are on same page. RFC patchset did that. But there are some other opens
on the RFC. So I tried to submit this part of change which is bug fix. The other
thing left in RFC is optimization (avoid split large folio if we can).

Regards
Yin, Fengwei

>
> Thanks,
> Ryan
>