On 2017/9/26 3:40, Steven Rostedt wrote:
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
Thanks for the reply! After the task_A sleep on cpu1, the try_to_wake_up(task_A) on cpu0 select a different cpu3, so it will grab the rq3 lock, not the rq1 lock. Thanks.
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>.
