Re: [PATCH RFC] sched: Revert delayed_put_task_struct() and fix use after free

From: Kirill Tkhai
Date: Wed Oct 15 2014 - 18:02:48 EST


On 16.10.2014 01:46, Kirill Tkhai wrote:
> Yeah, you're sure about initial patch. Thanks for signal explanation.
>
> On 15.10.2014 23:40, Oleg Nesterov wrote:
>> On 10/15, Oleg Nesterov wrote:
>>>
>>> On 10/15, Kirill Tkhai wrote:
>>>>
>>>> Regarding to scheduler this may be a reason of use-after-free.
>>>>
>>>> task_numa_compare() schedule()
>>>> rcu_read_lock() ...
>>>> cur = ACCESS_ONCE(dst_rq->curr) ...
>>>> ... rq->curr = next;
>>>> ... context_switch()
>>>> ... finish_task_switch()
>>>> ... put_task_struct()
>>>> ... __put_task_struct()
>>>> ... free_task_struct()
>>>> task_numa_assign() ...
>>>> get_task_struct() ...
>>>
>>> Agreed. I don't understand this code (will try to take another look later),
>>> but at first glance this looks wrong.
>>>
>>> At least the code like
>>>
>>> rcu_read_lock();
>>> get_task_struct(foreign_rq->curr);
>>> rcu_read_unlock();
>>>
>>> is certainly wrong. And _probably_ the problem should be fixed here. Perhaps
>>> we can add try_to_get_task_struct() which does atomic_inc_not_zero() ...
>>
>> Yes, but perhaps in this particular case another simple fix makes more
>> sense. The patch below needs a comment to explain that we check PF_EXITING
>> because:
>>
>> 1. It doesn't make sense to migrate the exiting task. Although perhaps
>> we could check ->mm == NULL instead.
>>
>> But let me repeat that I do not understand this code, I am not sure
>> we can equally treat is_idle_task() and PF_EXITING here...
>>
>> 2. If PF_EXITING is not set (or ->mm != NULL) then delayed_put_task_struct()
>> won't be called until we drop rcu_read_lock(), and thus get_task_struct()
>> is safe.
>>
>
> Cool! Elegant fix. We set PF_EXITING in exit_signals(), which is earlier
> than release_task() is called.
>
> Shouldn't we use smp_rmb/smp_wmb here?
>
>> And. it seems that there is another problem? Can't task_h_load(cur) race
>> with itself if 2 CPU's call task_numa_migrate() and inspect the same rq
>> in parallel? Again, I don't understand this code, but update_cfs_rq_h_load()
>> doesn't look "atomic". In fact I am not even sure about task_h_load(env->p),
>> p == current but we do not disable preemption.
>>
>> What do you think?
>
> We use it completely unlocked, so nothing good is here. Also we work
> with pointers.
>
> As I understand in update_cfs_rq_h_load() we go from bottom to top,
> and then from top to bottom. We set cfs_rq::h_load_next to be able
> to do top-bottom passage (top is a root of "tree").

> Yeah, this "way" may be overwritten by competitor. Also, task may change
> its cfs_rq.

Wrong, it's not a task... Brain is sleepy, it's better tomorrow.

>
>> --- x/kernel/sched/fair.c
>> +++ x/kernel/sched/fair.c
>> @@ -1165,7 +1165,7 @@ static void task_numa_compare(struct tas
>>
>> rcu_read_lock();
>> cur = ACCESS_ONCE(dst_rq->curr);
>> - if (cur->pid == 0) /* idle */
>> + if (is_idle_task(cur) || (curr->flags & PF_EXITING))
>> cur = NULL;
>>
>> /*
>>
>
> Looks like, we have to use the same fix for task_numa_group().
>
> grp = rcu_dereference(tsk->numa_group);
>
> Below we dereference grp->nr_tasks.
>
> Also, the same in rt.c and deadline.c, but we do no take second
> reference there. Wrong pointer dereference is not possible there,
> not so bad.
>
> Kirill
>

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