On 10/09/2012 07:35 PM, Michal Hocko wrote:
> On Tue 09-10-12 19:14:57, Glauber Costa wrote:
>> On 10/09/2012 07:08 PM, Michal Hocko wrote:
>>> As I have already mentioned in my previous feedback this is cetainly not
>>> atomic as you the lock protects only one group in the hierarchy. How is
>>> the return value from this function supposed to be used?
>>
>> So, I tried to make that clearer in the updated changelog.
>>
>> Only the value of the base memcg (the one passed to the function) is
>> returned, and it is atomic, in the sense that it has the same semantics
>> as the atomic variables: If 2 threads uncharge 4k each from a 8 k
>> counter, a subsequent read can return 0 for both. The return value here
>> will guarantee that only one sees the drop to 0.
>>
>> This is used in the patch "kmem_accounting lifecycle management" to be
>> sure that only one process will call mem_cgroup_put() in the memcg
>> structure.
>
> Yes, you are using res_counter_uncharge and its semantic makes sense.
> I was refering to res_counter_uncharge_until (you removed that context
> from my reply) because that one can race resulting that nobody sees 0
> even though that parents get down to 0 as a result:
> A
> |
> B
> / \
> C(x) D(y)
>
> D and C uncharge everything.
>
> CPU0 CPU1
> ret += uncharge(D) [0] ret += uncharge(C) [0]
> ret += uncharge(B) [x-from C]
> ret += uncharge(B) [0]
> ret += uncharge(A) [y-from D]
> ret += uncharge(A) [0]
>
> ret == x ret == y
>
Sorry Michal, I didn't realize you were talking about
res_counter_uncharge_until.
I don't really need res_counter_uncharge_until to return anything, so I
can just remove that if you prefer, keeping just the main
res_counter_uncharge.
However, I still can't make sense of your concern.
The return value will return the value of the counter passed as a
parameter to the function:
r = res_counter_uncharge_locked(c, val);
if (c == counter)
ret = r;
So when you call res_counter_uncharge_until(D, whatever, x), you will
see zero here as a result, and when you call
res_counter_uncharge_until(D, whatever, y) you will see 0 here as well.
A doesn't get involved with that.
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