Messages in this thread | | | Date | Thu, 16 Oct 2014 02:02:38 +0400 | From | Kirill Tkhai <> | Subject | Re: [PATCH RFC] sched: Revert delayed_put_task_struct() and fix use after free |
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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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