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.
Is my understanding correct? If not, could you please clarify on how to
get the lru items per node?
--
Qualcomm India Private Limited, on behalf of Qualcomm Innovation Center, Inc.
Qualcomm Innovation Center, Inc. is a member of Code Aurora Forum, a Linux Foundation Collaborative Project.
