On Wed, May 25, 2022 at 05:53:30PM +0800, Muchun Song wrote:
> On Tue, May 24, 2022 at 03:27:20PM -0400, Johannes Weiner wrote:
> > On Tue, May 24, 2022 at 02:05:43PM +0800, Muchun Song wrote:
> > > The diagram below shows how to make the folio lruvec lock safe when LRU
> > > pages are reparented.
> > >
> > > folio_lruvec_lock(folio)
> > > retry:
> > > lruvec = folio_lruvec(folio);
> > >
> > > // The folio is reparented at this time.
> > > spin_lock(&lruvec->lru_lock);
> > >
> > > if (unlikely(lruvec_memcg(lruvec) != folio_memcg(folio)))
> > > // Acquired the wrong lruvec lock and need to retry.
> > > // Because this folio is on the parent memcg lruvec list.
> > > goto retry;
> > >
> > > // If we reach here, it means that folio_memcg(folio) is stable.
> > >
> > > memcg_reparent_objcgs(memcg)
> > > // lruvec belongs to memcg and lruvec_parent belongs to parent memcg.
> > > spin_lock(&lruvec->lru_lock);
> > > spin_lock(&lruvec_parent->lru_lock);
> > >
> > > // Move all the pages from the lruvec list to the parent lruvec list.
> > >
> > > spin_unlock(&lruvec_parent->lru_lock);
> > > spin_unlock(&lruvec->lru_lock);
> > >
> > > After we acquire the lruvec lock, we need to check whether the folio is
> > > reparented. If so, we need to reacquire the new lruvec lock. On the
> > > routine of the LRU pages reparenting, we will also acquire the lruvec
> > > lock (will be implemented in the later patch). So folio_memcg() cannot
> > > be changed when we hold the lruvec lock.
> > >
> > > Since lruvec_memcg(lruvec) is always equal to folio_memcg(folio) after
> > > we hold the lruvec lock, lruvec_memcg_debug() check is pointless. So
> > > remove it.
> > >
> > > This is a preparation for reparenting the LRU pages.
> > >
> > > Signed-off-by: Muchun Song <songmuchun@xxxxxxxxxxxxx>
> >
> > This looks good to me. Just one question:
> >
> > > @@ -1230,10 +1213,23 @@ void lruvec_memcg_debug(struct lruvec *lruvec, struct folio *folio)
> > > */
> > > struct lruvec *folio_lruvec_lock(struct folio *folio)
> > > {
> > > - struct lruvec *lruvec = folio_lruvec(folio);
> > > + struct lruvec *lruvec;
> > >
> > > + rcu_read_lock();
> > > +retry:
> > > + lruvec = folio_lruvec(folio);
> > > spin_lock(&lruvec->lru_lock);
> > > - lruvec_memcg_debug(lruvec, folio);
> > > +
> > > + if (unlikely(lruvec_memcg(lruvec) != folio_memcg(folio))) {
> > > + spin_unlock(&lruvec->lru_lock);
> > > + goto retry;
> > > + }
> > > +
> > > + /*
> > > + * Preemption is disabled in the internal of spin_lock, which can serve
> > > + * as RCU read-side critical sections.
> > > + */
> > > + rcu_read_unlock();
> >
> > The code looks right to me, but I don't understand the comment: why do
> > we care that the rcu read-side continues? With the lru_lock held,
> > reparenting is on hold and the lruvec cannot be rcu-freed anyway, no?
> >
>
> Right. We could hold rcu read lock until end of reparting. So you mean
> we do rcu_read_unlock in folio_lruvec_lock()?
The comment seems to suggest that disabling preemption is what keeps
the lruvec alive. But it's the lru_lock that keeps it alive. The
cgroup destruction path tries to take the lru_lock long before it even
gets to synchronize_rcu(). Once you hold the lru_lock, having an
implied read-side critical section as well doesn't seem to matter.
Should the comment be deleted?
