Messages in this thread | | | Date | Thu, 16 Oct 2014 01:46:07 +0400 | From | Kirill Tkhai <> | Subject | Re: [PATCH RFC] sched: Revert delayed_put_task_struct() and fix use after free |
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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.
> --- 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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