Re: [PATCH v2] memcg: Only record foreign writebacks with dirty pages when memcg is not disabled

From: Michal Hocko
Date: Tue Sep 24 2019 - 09:46:09 EST


On Tue 24-09-19 21:43:52, Baoquan He wrote:
> On 09/24/19 at 03:16pm, Michal Hocko wrote:
> > On Tue 24-09-19 21:04:58, Baoquan He wrote:
> > > On 09/24/19 at 02:27pm, Michal Hocko wrote:
> > > > On Tue 24-09-19 19:11:51, Baoquan He wrote:
> > > > > diff --git a/mm/memcontrol.c b/mm/memcontrol.c
> > > > > index f3c15bb07cce..84e3fdb1ccb4 100644
> > > > > --- a/mm/memcontrol.c
> > > > > +++ b/mm/memcontrol.c
> > > > > @@ -4317,6 +4317,9 @@ void mem_cgroup_track_foreign_dirty_slowpath(struct page *page,
> > > > >
> > > > > trace_track_foreign_dirty(page, wb);
> > > > >
> > > > > + if (mem_cgroup_disabled())
> > > > > + return;
> > > > > +
> > > >
> > > > This doesn't seem correct. We shouldn't even enter the slowpath with
> > > > memcg disabled AFAIC. The check should be done at mem_cgroup_track_foreign_dirty
> > > > level.
> > >
> > > You mean the way in v1 patch, right? It's also fine to me.
> > >
> > > I am worried about the case that memcg is enabled, the checking by
> > > calling mem_cgroup_disabled() will lower efficiency.
> >
> > This is hidden by a static branch so I wouldn't really be worried about
> > the overhead.
> >
> > > And it entering
> > > into mem_cgroup_track_foreign_dirty_slowpath() should be a rare event.
> >
> > But &page->mem_cgroup->css != wb->memcg_css doesn't make any sense when
> > memcg is disabled, right?
>
> Yeah, I think so. Make it like below?

Or just put it on its own line to make the code more readable.

> @@ -1261,7 +1261,8 @@ void mem_cgroup_track_foreign_dirty_slowpath(struct page *page,
> static inline void mem_cgroup_track_foreign_dirty(struct page *page,
> struct bdi_writeback *wb)
> {
> - if (unlikely(&page->mem_cgroup->css != wb->memcg_css))
> + if (!mem_cgroup_disabled() &&
> + unlikely(&page->mem_cgroup->css != wb->memcg_css))
> mem_cgroup_track_foreign_dirty_slowpath(page, wb);
> }

--
Michal Hocko
SUSE Labs