linux-kernel - Re:[PATCH] sched/numa: fix unsafe get_task_struct() in task_numa

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Message-Id: <4594081413623803@web19g.yandex.ru>
Date:	Sat, 18 Oct 2014 13:16:43 +0400
From:	Kirill Tkhai <tkhai@...dex.ru>
To:	Kirill Tkhai <tkhai@...dex.ru>
Cc:	linux-kernel@...r.kernel.org, Ingo Molnar <mingo@...hat.com>,
	Vladimir Davydov <vdavydov@...allels.com>,
	Oleg Nesterov <oleg@...hat.com>,
	Kirill Tkhai <ktkhai@...allels.com>,
	Peter Zijlstra <peterz@...radead.org>
Subject: Re:[PATCH] sched/numa: fix unsafe get_task_struct() in task_numa_assign()

And smp_rmb() beetween ifs which is pairs with rq unlocking

> 18.10.2014, 12:15, "Kirill Tkhai" <tkhai@...dex.ru>:
> 
>> 18.10.2014, 01:40, "Oleg Nesterov" <oleg@...hat.com>:
>>
>>> The lockless get_task_struct(tsk) is only safe if tsk == current
>>> and didn't pass exit_notify(), or if this tsk was found on a rcu
>>> protected list (say, for_each_process() or find_task_by_vpid()).
>>> IOW, it is only safe if release_task() was not called before we
>>> take rcu_read_lock(), in this case we can rely on the fact that
>>> delayed_put_pid() can not drop the (potentially) last reference
>>> until rcu_read_unlock().
>>>
>>> And as Kirill pointed out task_numa_compare()->task_numa_assign()
>>> path does get_task_struct(dst_rq->curr) and this is not safe. The
>>> task_struct itself can't go away, but rcu_read_lock() can't save
>>> us from the final put_task_struct() in finish_task_switch(); this
>>> reference goes away without rcu gp.
>>>
>>> Reported-by: Kirill Tkhai <ktkhai@...allels.com>
>>> Signed-off-by: Oleg Nesterov <oleg@...hat.com>
>>> ---
>>> kernel/sched/fair.c | 8 +++++++-
>>> 1 files changed, 7 insertions(+), 1 deletions(-)
>>>
>>> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
>>> index 0090e8c..52049b9 100644
>>> --- a/kernel/sched/fair.c
>>> +++ b/kernel/sched/fair.c
>>> @@ -1158,7 +1158,13 @@ static void task_numa_compare(struct task_numa_env *env,
>>>
>>> rcu_read_lock();
>>> cur = ACCESS_ONCE(dst_rq->curr);
>>> - if (cur->pid == 0) /* idle */
>>> + /*
>>> + * No need to move the exiting task, and this ensures that ->curr
>>> + * wasn't reaped and thus get_task_struct() in task_numa_assign()
>>> + * is safe; note that rcu_read_lock() can't protect from the final
>>> + * put_task_struct() after the last schedule().
>>> + */
>>> + if (is_idle_task(cur) || (cur->flags & PF_EXITING))
>>> cur = NULL;
>>>
>>> /*
>>
>> Oleg, I've looked once again, and now it's not good for me.
>> Where is the guarantee this memory hasn't been allocated again?
>> If so, PF_EXITING is not of the task we are interesting, but it's
>> not a task's even.
>>
>> rcu_read_lock() ... ...
>> cur = ACCESS_ONCE(dst_rq->curr); ... ...
>> <interrupt> rq->curr = next; ...
>> <interrupt> put_prev_task() ...
>> <interrupt> __put_prev_task ...
>> <interrupt> kmem_cache_free() ...
>> <interrupt> ... <alocated again>
>> <interrupt> ... memset(, 0, )
>> <interrupt> ... ...
>> if (cur->flags & PF_EXITING) ... ...
>> <no> ... ...
>> get_task_struct() ... ...
> 
> How about this?
> 
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index b78280c..d46427e 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -1165,7 +1165,21 @@ static void task_numa_compare(struct task_numa_env *env,
> 
> rcu_read_lock();
> cur = ACCESS_ONCE(dst_rq->curr);
> - if (cur->pid == 0) /* idle */
> + /*
> + * No need to move the exiting task, and this ensures that ->curr
> + * wasn't reaped and thus get_task_struct() in task_numa_assign()
> + * is safe; note that rcu_read_lock() can't protect from the final
> + * put_task_struct() after the last schedule().
> + */
> + if (is_idle_task(cur) || (cur->flags & PF_EXITING))
> + cur = NULL;
> + /*
> + * Check once again to be sure curr is still on dst_rq. Even if
> + * it points on a new task, which is using the memory of freed
> + * cur, it's OK, because we've locked RCU before
> + * delayed_put_task_struct() callback is called to put its struct.
> + */
> + if (cur != ACCESS_ONCE(dst_rq->curr))
> cur = NULL;
> 
> /*
> --
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