[Devel] [PATCH 6/6] memcg, slab: RCU protect memcg_params for root caches

[Vladimir Davydov]

On 12/19/2013 01:43 PM, Michal Hocko wrote:
> On Thu 19-12-13 13:36:42, Vladimir Davydov wrote:
>> On 12/19/2013 01:28 PM, Michal Hocko wrote:
>>> On Wed 18-12-13 17:16:57, Vladimir Davydov wrote:
> [...]
>>>> diff --git a/mm/slab.h b/mm/slab.h
>>>> index 1d8b53f..53b81a9 100644
>>>> --- a/mm/slab.h
>>>> +++ b/mm/slab.h
>>>> @@ -164,10 +164,16 @@ static inline struct kmem_cache *
>>>>  cache_from_memcg_idx(struct kmem_cache *s, int idx)
>>>>  {
>>>>  	struct kmem_cache *cachep;
>>>> +	struct memcg_cache_params *params;
>>>>  
>>>>  	if (!s->memcg_params)
>>>>  		return NULL;
>>>> -	cachep = s->memcg_params->memcg_caches[idx];
>>>> +
>>>> +	rcu_read_lock();
>>>> +	params = rcu_dereference(s->memcg_params);
>>>> +	cachep = params->memcg_caches[idx];
>>>> +	rcu_read_unlock();
>>>> +
>>> Consumer has to be covered by the same rcu section otherwise
>>> memcg_params might be freed right after rcu unlock here.
>> No. We protect only accesses to kmem_cache::memcg_params, which can
>> potentially be relocated for root caches.
> Hmm, ok. So memcg_params might change (a new memcg is accounted) but
> pointers at idx will be same, right?

Yes, that's a classical Read-Copy-Update :-)

>
>> But as soon as we get the
>> pointer to a kmem_cache from this array, we can freely dereference it,
>> because the cache cannot be freed when we use it. This is, because we
>> access a kmem_cache either under the slab_mutex or
>> memcg->slab_caches_mutex, or when we allocate/free from it. While doing
>> the latter, the cache can't go away, it would be a bug. IMO.
> That expects that cache_from_memcg_idx is always called with slab_mutex
> or slab_caches_mutex held, right? Please document it.

Yeah, you're right, this longs for a documentation. I'm going to check
this code a bit more and try to write a good comment about it (although
I'm rather poor at writing comments :-( )

Thanks.