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Subject: Re: [PATCH 06/10] list_lru: Pass dst_memcg argument to
 memcg_drain_list_lru_node()
From:   Kirill Tkhai <ktkhai@virtuozzo.com>
To:     Vladimir Davydov <vdavydov.dev@gmail.com>
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Date:   Wed, 28 Mar 2018 17:49:34 +0300
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On 26.03.2018 18:30, Kirill Tkhai wrote:
> On 24.03.2018 22:32, Vladimir Davydov wrote:
>> On Wed, Mar 21, 2018 at 04:22:10PM +0300, Kirill Tkhai wrote:
>>> This is just refactoring to allow next patches to have
>>> dst_memcg pointer in memcg_drain_list_lru_node().
>>>
>>> Signed-off-by: Kirill Tkhai <ktkhai@virtuozzo.com>
>>> ---
>>>  include/linux/list_lru.h |    2 +-
>>>  mm/list_lru.c            |   11 ++++++-----
>>>  mm/memcontrol.c          |    2 +-
>>>  3 files changed, 8 insertions(+), 7 deletions(-)
>>>
>>> diff --git a/include/linux/list_lru.h b/include/linux/list_lru.h
>>> index ce1d010cd3fa..50cf8c61c609 100644
>>> --- a/include/linux/list_lru.h
>>> +++ b/include/linux/list_lru.h
>>> @@ -66,7 +66,7 @@ int __list_lru_init(struct list_lru *lru, bool memcg_aware,
>>>  #define list_lru_init_memcg(lru)	__list_lru_init((lru), true, NULL)
>>>  
>>>  int memcg_update_all_list_lrus(int num_memcgs);
>>> -void memcg_drain_all_list_lrus(int src_idx, int dst_idx);
>>> +void memcg_drain_all_list_lrus(int src_idx, struct mem_cgroup *dst_memcg);
>>
>> Please, for consistency pass the source cgroup as a pointer as well.
> 
> Ok

Hm. But we call it from memcg_offline_kmem() after cgroup's kmemcg_id is set
to parent memcg's kmemcg_id:

        rcu_read_lock(); /* can be called from css_free w/o cgroup_mutex */
        css_for_each_descendant_pre(css, &memcg->css) {
                child = mem_cgroup_from_css(css);
                BUG_ON(child->kmemcg_id != kmemcg_id);
                child->kmemcg_id = parent->kmemcg_id;
                if (!memcg->use_hierarchy)
                        break;
        }
        rcu_read_unlock();

        memcg_drain_all_list_lrus(kmemcg_id, parent);

It does not seem we may pass memcg to memcg_drain_all_list_lrus()
or change the logic or order. What do you think?

Kirill