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.
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?
Oleg.
--- 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;
/*
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