Re: [PATCH v4 03/11] mm: memcontrol: make lruvec lock safe when LRU pages are reparented

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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()?

Thanks.



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