On 13.04.2018 14:38, Michal Hocko wrote: > On Fri 13-04-18 14:29:11, Kirill Tkhai wrote: >> On 13.04.2018 14:20, Michal Hocko wrote: >>> On Fri 13-04-18 14:06:40, Kirill Tkhai wrote: >>>> On 13.04.2018 14:02, Michal Hocko wrote: >>>>> On Fri 13-04-18 12:35:22, Kirill Tkhai wrote: >>>>>> On 13.04.2018 11:55, Michal Hocko wrote: >>>>>>> On Thu 12-04-18 17:52:04, Kirill Tkhai wrote: >>>>>>> [...] >>>>>>>> @@ -4471,6 +4477,7 @@ mem_cgroup_css_alloc(struct cgroup_subsys_state >>>>>>>> *parent_css) >>>>>>>> >>>>>>>> return &memcg->css; >>>>>>>> fail: >>>>>>>> + mem_cgroup_id_remove(memcg); >>>>>>>> mem_cgroup_free(memcg); >>>>>>>> return ERR_PTR(-ENOMEM); >>>>>>>> } >>>>>>> >>>>>>> The only path which jumps to fail: here (in the current mmotm tree) is >>>>>>> error = memcg_online_kmem(memcg); >>>>>>> if (error) >>>>>>> goto fail; >>>>>>> >>>>>>> AFAICS and the only failure path in memcg_online_kmem >>>>>>> memcg_id = memcg_alloc_cache_id(); >>>>>>> if (memcg_id < 0) >>>>>>> return memcg_id; >>>>>>> >>>>>>> I am not entirely clear on memcg_alloc_cache_id but it seems we do clean >>>>>>> up properly. Or am I missing something? >>>>>> >>>>>> memcg_alloc_cache_id() may allocate a lot of memory, in case of the >>>>>> system reached >>>>>> memcg_nr_cache_ids cgroups. In this case it iterates over all LRU lists, >>>>>> and double >>>>>> size of every of them. In case of memory pressure it can fail. If this >>>>>> occurs, >>>>>> mem_cgroup::id is not unhashed from IDR and we leak this id. >>>>> >>>>> OK, my bad I was looking at the bad code path. So you want to clean up >>>>> after mem_cgroup_alloc not memcg_online_kmem. Now it makes much more >>>>> sense. Sorry for the confusion on my end. >>>>> >>>>> Anyway, shouldn't we do the thing in mem_cgroup_free() to be symmetric >>>>> to mem_cgroup_alloc? >>>> >>>> We can't, since it's called from mem_cgroup_css_free(), which doesn't have >>>> a deal >>>> with idr freeing. All the asymmetry, we see, is because of the trick to >>>> unhash ID >>>> earlier, then from mem_cgroup_css_free(). >>> >>> Are you sure. It's been some time since I've looked at the quite complex >>> cgroup tear down code but from what I remember, css_free is called on >>> the css release (aka when the reference count drops to zero). >>> mem_cgroup_id_put_many >>> seems to unpin the css reference so we should have idr_remove by the >>> time when css_free is called. Or am I still wrong and should go over the >>> brain hurting cgroup removal code again? >> >> mem_cgroup_id_put_many() unpins css, but this may be not the last reference >> to the css. >> Thus, we release ID earlier, then all references to css are freed. > > Right and so what. If we have released the idr then we are not going to > do that again in css_free. That is why we have that memcg->id.id > 0 > check before idr_remove and memcg->id.id = 0 for the last memcg ref. > count. So again, why cannot we do the clean up in mem_cgroup_free and > have a less confusing code? Or am I just not getting your point and > being dense here?
We can, but mem_cgroup_free() called from mem_cgroup_css_alloc() is unlikely case. The likely case is mem_cgroup_free() is called from mem_cgroup_css_free(), where this idr manipulations will be a noop. Noop in likely case looks more confusing for me. Less confusing will be to move memcg->id.id = idr_alloc(&mem_cgroup_idr, NULL, 1, MEM_CGROUP_ID_MAX, GFP_KERNEL); into mem_cgroup_css_alloc(). How are you think about this? Kirill