sched/numa: Fix unsafe get_task_struct() in task_numa_assign()
Unlocked access to dst_rq->curr in task_numa_compare() is racy.
If curr task is exiting this may be a reason of use-after-free:
task_numa_compare() do_exit()
... current->flags |= PF_EXITING;
... release_task()
... ~~delayed_put_task_struct()~~
... 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() ...
As noted by Oleg:
<<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>>
The patch provides simple check of PF_EXITING flag. If it's not set,
this guarantees that call_rcu() of delayed_put_task_struct() callback
hasn't happened yet, so we can safely do get_task_struct() in
task_numa_assign().
Locked dst_rq->lock protects from concurrency with the last schedule().
Reusing or unmapping of cur's memory may happen without it.
Suggested-by: Oleg Nesterov <oleg@redhat.com>
Signed-off-by: Kirill Tkhai <ktkhai@parallels.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Link: http://lkml.kernel.org/r/1413962231.19914.130.camel@tkhai
Signed-off-by: Ingo Molnar <mingo@kernel.org>