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