linux-kernel - Re: [PATCH v7 3/4] mm,page_owner: Display all stacks and their count

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Date: Sat, 10 Feb 2024 08:52:25 +0100
From: Marco Elver <elver@...gle.com>
To: Oscar Salvador <osalvador@...e.de>
Cc: Andrew Morton <akpm@...ux-foundation.org>, linux-kernel@...r.kernel.org, 
	linux-mm@...ck.org, Michal Hocko <mhocko@...e.com>, Vlastimil Babka <vbabka@...e.cz>, 
	Andrey Konovalov <andreyknvl@...il.com>, Alexander Potapenko <glider@...gle.com>
Subject: Re: [PATCH v7 3/4] mm,page_owner: Display all stacks and their count

On Sat, 10 Feb 2024 at 00:13, Oscar Salvador <osalvador@...e.de> wrote:
>
> On Fri, Feb 09, 2024 at 10:52:48PM +0100, Oscar Salvador wrote:
> > Thinking about it some more, I think I made a mistake:
> >
> > I am walking all buckets, and within those buckets there are not only
> > page_owner stack_records, which means that I could return a stack_record
> > from e.g: KASAN (which I think can evict stack_records) and then
> > everything goes off the rails.
> > Which means I cannot walk the buckets like that.
> >
> > Actually, I think that having something like the following
> >
> >  struct list_stack_records {
> >       struct stack_record *stack;
> >       struct list_stack_records *next;
> >  }
>
> Or, I could use the extra_bits field from handle_parts to flag that
> when a depot_stack_handle_t is used by page_owner.
>
> Then __stack_depot_get_next_stack_record() would check whether
> a stack_record->handle.extra_bits has the page_owner bit, and only
> return those stacks that have such bit.
> This would solve the problem of returning a potentially evictable stack
> , only by returning page_owner's stack_records, and I would not have
> to maintain my own list.
>
> I yet have to see how that would look like, but sounds promising.
> Do you think that is feasible Marco?

The extra bits are used by KMSAN, and might conflict if enabled at the
same time. I think the safest option is to keep your own list. I think
that will also be more performant if there are other stackdepot users
because you do not have to traverse any of the other entries.