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    SubjectRe: [PATCH v4 03/11] mm: memcontrol: make lruvec lock safe when LRU pages are reparented
    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@bytedance.com>
    >
    > 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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