On Mon, 11 Sep 2017 14:51:49 +0800
Zhou Chengming <zhouchengmi...@huawei.com> wrote:
> push_rt_task() pick the first pushable task and find an eligible
> lowest_rq, then double_lock_balance(rq, lowest_rq). So if
> double_lock_balance() unlock the rq (when double_lock_balance() return 1),
> we have to check if this task is still on the rq.
>
> The problem is that the check conditions are not sufficient:
>
> if (unlikely(task_rq(task) != rq ||
> !cpumask_test_cpu(lowest_rq->cpu, &task->cpus_allowed) ||
> task_running(rq, task) ||
> !rt_task(task) ||
> !task_on_rq_queued(task))) {
>
> cpu2 cpu1 cpu0
> push_rt_task(rq1)
> pick task_A on rq1
> find rq0
> double_lock_balance(rq1, rq0)
> unlock(rq1)
> rq1 __schedule
> pick task_A run
> task_A sleep (dequeued)
> lock(rq0)
> lock(rq1)
> do_above_check(task_A)
> task_rq(task_A) == rq1
> cpus_allowed unchanged
> task_running == false
> rt_task(task_A) == true
> try_to_wake_up(task_A)
> select_cpu = cpu3
> enqueue(rq3, task_A)
How can this happen? The try_to_wake_up(task_A) needs to grab the rq
that task A is on, and we have that rq lock.
/me confused.
-- Steve
> task_A->on_rq = 1
> task_on_rq_queued(task_A)
> above_check passed, return rq0
> ...
> migrate task_A from rq1 to rq0
>
> So we can't rely on these checks of task_A to make sure the task_A is
> still on the rq1, even though we hold the rq1->lock. This patch will
> repick the first pushable task to be sure the task is still on the rq.
>
> Signed-off-by: Zhou Chengming <zhouchengmi...@huawei.com>
>
