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SubjectRe: [mm] [PATCH 3/4] Memory cgroup hierarchical reclaim
KAMEZAWA Hiroyuki wrote:
> On Sun, 02 Nov 2008 11:14:48 +0530
> Balbir Singh <balbir@linux.vnet.ibm.com> wrote:
>
>> KAMEZAWA Hiroyuki wrote:
>>> On Sun, 02 Nov 2008 00:18:49 +0530
>>> Balbir Singh <balbir@linux.vnet.ibm.com> wrote:
>>>
>>>> This patch introduces hierarchical reclaim. When an ancestor goes over its
>>>> limit, the charging routine points to the parent that is above its limit.
>>>> The reclaim process then starts from the last scanned child of the ancestor
>>>> and reclaims until the ancestor goes below its limit.
>>>>
>>>> Signed-off-by: Balbir Singh <balbir@linux.vnet.ibm.com>
>>>> ---
>>>>
>>>> mm/memcontrol.c | 153 +++++++++++++++++++++++++++++++++++++++++++++++---------
>>>> 1 file changed, 129 insertions(+), 24 deletions(-)
>>>>
>>>> diff -puN mm/memcontrol.c~memcg-hierarchical-reclaim mm/memcontrol.c
>>>> --- linux-2.6.28-rc2/mm/memcontrol.c~memcg-hierarchical-reclaim 2008-11-02 00:14:59.000000000 +0530
>>>> +++ linux-2.6.28-rc2-balbir/mm/memcontrol.c 2008-11-02 00:14:59.000000000 +0530
>>>> @@ -132,6 +132,11 @@ struct mem_cgroup {
>>>> * statistics.
>>>> */
>>>> struct mem_cgroup_stat stat;
>>>> + /*
>>>> + * While reclaiming in a hiearchy, we cache the last child we
>>>> + * reclaimed from.
>>>> + */
>>>> + struct mem_cgroup *last_scanned_child;
>>>> };
>>>> static struct mem_cgroup init_mem_cgroup;
>>>>
>>>> @@ -467,6 +472,125 @@ unsigned long mem_cgroup_isolate_pages(u
>>>> return nr_taken;
>>>> }
>>>>
>>>> +static struct mem_cgroup *
>>>> +mem_cgroup_from_res_counter(struct res_counter *counter)
>>>> +{
>>>> + return container_of(counter, struct mem_cgroup, res);
>>>> +}
>>>> +
>>>> +/*
>>>> + * Dance down the hierarchy if needed to reclaim memory. We remember the
>>>> + * last child we reclaimed from, so that we don't end up penalizing
>>>> + * one child extensively based on its position in the children list
>>>> + */
>>>> +static int
>>>> +mem_cgroup_hierarchical_reclaim(struct mem_cgroup *mem, gfp_t gfp_mask)
>>>> +{
>>>> + struct cgroup *cg, *cg_current, *cgroup;
>>>> + struct mem_cgroup *mem_child;
>>>> + int ret = 0;
>>>> +
>>>> + if (try_to_free_mem_cgroup_pages(mem, gfp_mask))
>>>> + return -ENOMEM;
>>>> +
>>>> + /*
>>>> + * try_to_free_mem_cgroup_pages() might not give us a full
>>>> + * picture of reclaim. Some pages are reclaimed and might be
>>>> + * moved to swap cache or just unmapped from the cgroup.
>>>> + * Check the limit again to see if the reclaim reduced the
>>>> + * current usage of the cgroup before giving up
>>>> + */
>>>> + if (res_counter_check_under_limit(&mem->res))
>>>> + return 0;
>>>> +
>>>> + /*
>>>> + * Scan all children under the mem_cgroup mem
>>>> + */
>>>> + if (!mem->last_scanned_child)
>>>> + cgroup = list_first_entry(&mem->css.cgroup->children,
>>>> + struct cgroup, sibling);
>>>> + else
>>>> + cgroup = mem->last_scanned_child->css.cgroup;
>>>> +
>>>> + cg_current = cgroup;
>>>> +
>>>> + /*
>>>> + * We iterate twice, one of it is fundamental list issue, where
>>>> + * the elements are inserted using list_add and hence the list
>>>> + * behaves like a stack and list_for_entry_safe_from() stops
>>>> + * after seeing the first child. The two loops help us work
>>>> + * independently of the insertion and it helps us get a full pass at
>>>> + * scanning all list entries for reclaim
>>>> + */
>>>> + list_for_each_entry_safe_from(cgroup, cg, &cg_current->parent->children,
>>>> + sibling) {
>>>> + mem_child = mem_cgroup_from_cont(cgroup);
>>>> +
>>>> + /*
>>>> + * Move beyond last scanned child
>>>> + */
>>>> + if (mem_child == mem->last_scanned_child)
>>>> + continue;
>>>> +
>>>> + ret = try_to_free_mem_cgroup_pages(mem_child, gfp_mask);
>>>> + mem->last_scanned_child = mem_child;
>>>> +
>>>> + if (res_counter_check_under_limit(&mem->res)) {
>>>> + ret = 0;
>>>> + goto done;
>>>> + }
>>>> + }
>>> Is this safe against cgroup create/remove ? cgroup_mutex is held ?
>> Yes, I thought about it, but with the setup, each parent will be busy since they
>> have children and hence cannot be removed. The leaf child itself has tasks, so
>> it cannot be removed. IOW, it should be safe against removal.
>>
> I'm sorry if I misunderstand something. could you explain folloing ?
>
> In following tree,
>
> level-1
> - level-2
> - level-3
> - level-4
> level-1's usage = level-1 + level-2 + level-3 + level-4
> level-2's usage = level-2 + level-3 + level-4
> level-3's usage = level-3 + level-4
> level-4's usage = level-4
>
> Assume that a task in level-2 hits its limit. It has to reclaim memory from
> level-2 and level-3, level-4.
>
> How can we guarantee level-4 has a task in this case ?

Good question. If there is no task, the LRU's will be empty and reclaim will
return. We could also add other checks if needed.

--
Balbir


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