On Thu, Jun 22, 2017 at 10:01:39PM +0530, Sahitya Tummala wrote:
>
>
> On 6/21/2017 10:01 PM, Vladimir Davydov wrote:
> >
> >>index cddf397..c8ca150 100644
> >>--- a/fs/dcache.c
> >>+++ b/fs/dcache.c
> >>@@ -1133,10 +1133,11 @@ void shrink_dcache_sb(struct super_block *sb)
> >> LIST_HEAD(dispose);
> >> freed = list_lru_walk(&sb->s_dentry_lru,
> >>- dentry_lru_isolate_shrink, &dispose, UINT_MAX);
> >>+ dentry_lru_isolate_shrink, &dispose, 1024);
> >> this_cpu_sub(nr_dentry_unused, freed);
> >> shrink_dentry_list(&dispose);
> >>+ cond_resched();
> >> } while (freed > 0);
> >In an extreme case, a single invocation of list_lru_walk() can skip all
> >1024 dentries, in which case 'freed' will be 0 forcing us to break the
> >loop prematurely. I think we should loop until there's at least one
> >dentry left on the LRU, i.e.
> >
> > while (list_lru_count(&sb->s_dentry_lru) > 0)
> >
> >However, even that wouldn't be quite correct, because list_lru_count()
> >iterates over all memory cgroups to sum list_lru_one->nr_items, which
> >can race with memcg offlining code migrating dentries off a dead cgroup
> >(see memcg_drain_all_list_lrus()). So it looks like to make this check
> >race-free, we need to account the number of entries on the LRU not only
> >per memcg, but also per node, i.e. add list_lru_node->nr_items.
> >Fortunately, list_lru entries can't be migrated between NUMA nodes.
> It looks like list_lru_count() is iterating per node before iterating over
> all memory
> cgroups as below -
>
> unsigned long list_lru_count_node(struct list_lru *lru, int nid)
> {
> long count = 0;
> int memcg_idx;
>
> count += __list_lru_count_one(lru, nid, -1);
> if (list_lru_memcg_aware(lru)) {
> for_each_memcg_cache_index(memcg_idx)
> count += __list_lru_count_one(lru, nid, memcg_idx);
> }
> return count;
> }
>
> The first call to __list_lru_count_one() is iterating all the items per node
> i.e, nlru->lru->nr_items.
lru->node[nid].lru.nr_items returned by __list_lru_count_one(lru, nid, -1)
only counts items accounted to the root cgroup, not the total number of
entries on the node.
> Is my understanding correct? If not, could you please clarify on how to get
> the lru items per node?
What I mean is iterating over list_lru_node->memcg_lrus to count the
number of entries on the node is racy. For example, suppose you have
three cgroups with the following values of list_lru_one->nr_items:
0 0 10
While list_lru_count_node() is at #1, cgroup #2 is offlined and its
list_lru_one is drained, i.e. its entries are migrated to the parent
cgroup, which happens to be #0, i.e. we see the following picture:
10 0 0
^^^
memcg_ids points here in list_lru_count_node()
Then the count returned by list_lru_count_node() will be 0, although
there are still 10 entries on the list.
To avoid this race, we could keep list_lru_node->lock locked while
walking over list_lru_node->memcg_lrus, but that's too heavy. I'd prefer
adding list_lru_node->nr_count which would be equal to the total number
of list_lru entries on the node, i.e. sum of list_lru_node->lru.nr_lrus
and list_lru_node->memcg_lrus->lru[]->nr_items.
