On 12/09/2013 07:22 PM, Michal Hocko wrote:
> On Wed 04-12-13 15:56:51, Vladimir Davydov wrote:
>> On 12/04/2013 02:08 PM, Glauber Costa wrote:
>>>>> Could you do something clever with just one flag? Probably yes. But I
>>>>> doubt it would
>>>>> be that much cleaner, this is just the way that patching sites work.
>>>> Thank you for spending your time to listen to me.
>>>>
>>> Don't worry! I thank you for carrying this forward.
>>>
>>>> Let me try to explain what is bothering me.
>>>>
>>>> We have two state bits for each memcg, 'active' and 'activated'. There
>>>> are two scenarios where the bits can be modified:
>>>>
>>>> 1) The kmem limit is set on a memcg for the first time -
>>>> memcg_update_kmem_limit(). Here we call memcg_update_cache_sizes(),
>>>> which sets the 'activated' bit on success, then update static branching,
>>>> then set the 'active' bit. All three actions are done atomically in
>>>> respect to other tasks setting the limit due to the set_limit_mutex.
>>>> After both bits are set, they never get cleared for the memcg.
>>>>
>>> So far so good. But again, note how you yourself describe it:
>>> the cations are done atomically *in respect to other tasks setting the
>>> limit*
>>>
>>> But there are also tasks that are running its courses naturally and
>>> just allocating
>>> memory. For those, some call sites will be on, some will be off. We need to
>>> make
>>> sure that *none* of them uses the patched site until *all* of them are
>>> patched.
>>> This has nothing to do with updates, this is all about the readers.
>>>
>>>> 2) When a subgroup of a kmem-active cgroup is created -
>>>> memcg_propagate_kmem(). Here we copy kmem_account_flags from the parent,
>>>> then increase static branching refcounter, then call
>>>> memcg_update_cache_sizes() for the new memcg, which may clear the
>>>> 'activated' bit on failure. After successful execution, the state bits
>>>> never get cleared for the new memcg.
>>>>
>>>> In scenario 2 there is no need bothering about the flags setting order,
>>>> because we don't have any tasks in the cgroup yet - the tasks can be
>>>> moved in only after css_online finishes when we have both of the bits
>>>> set and the static branching enabled. Actually, we already do not bother
>>>> about it, because we have both bits set before the cgroup is fully
>>>> initialized (memcg_update_cache_sizes() is called).
>>>>
>>> Yes, after the first cgroup is set none of that matters. But it is just
>>> easier
>>> and less error prone just to follow the same path every time. As I have
>>> said,
>>> if you can come up with a more clever way to deal with the problem above
>>> that doesn't involve the double flag - and you can prove it works - I
>>> am definitely
>>> fine with it. But this is subtle code, and in the past - Michal can
>>> attest this - we've
>>> changed this being sure it would work just to see it explode in our faces.
>>>
>>> So although I am willing to review every patch for correctness on that
>>> front (I never
>>> said I liked the 2-flags scheme...), unless you have a bug or real
>>> problem on it,
>>> I would advise against changing it if its just to make it more readable.
>>>
>>> But again, don't take me too seriously on this. If you and Michal think you
>>> can
>>> come up with something better, I'm all for it.
>> All right, I finally get you :-)
>>
>> Although I still don't think we need the second flag, I now understand
>> that it's better not to change the code that works fine especially the
>> change does not make it neither more readable nor more effective. Since
>> I can be mistaken about the flags usage (it's by far not unlikely), it's
>> better to leave it as is rather than being at risk of catching spurious
>> hangs that might be caused by this modification.
>>
>> Thanks for the detailed explanation!
> It would be really great if we could push some of that into the
> comments, please?
>
> Anyway, reading this thread again, I guess I finally got what you meant
> Vladimir.
> You are basically saying that the two stage enabling can be done
> by static_key_slow_inc in the first step and memcg_kmem_set_active
> in the second step without an additional flag.
> Assuming that the writers cannot race (they cannot currently because
> they are linearized by set_limit_mutex and memcg_create_mutex) and
> readers (charging paths) are _always_ checking the static key before
> checking active flags?
Right. There is no point in checking the static key after checking
active flags, because the benefit of using static branching would
disappear then. So IMHO the only thing we should bother is that the
static key refcounter is incremented *before* the active bit is set.
That assures all static branches have been patched if a charge path
succeeds, because a charge path cannot succeed if the active bit is not
set. That said we won't skip a commit or uncharge after a charge due to
an unpatched static branch. That's why I think the 'active' bit is enough.
Currently we have two flags 'activated' and 'active', and their usage
looks strange to me. Throughout the code we only have the following checks:
test_bit('active', state_mask)
test_bit('active', state_mask)&&test_bit('activated', state_mask)
Since 'active' bit is always set after 'activated' and none of them gets
cleared, the latter check is equivalent to the former.
Since we never issue a check like this:
test_bit('activated', state_mask)
we never actually check the 'activated' bit and do not need it - ???
Thanks.
> I guess this should work. But it would require a deep audit that the
> above is correct in all places. For example we do not bother to check
> static key during offline/free paths. I guess it should be harmless as
> is but who knows...
>
> I would rather see more detailed description of the current state first.
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