Re: [PATCH] sched/rt.c: pick and check task if double_lock_balance() unlock the rq

From: zhouchengming
Date: Mon Sep 25 2017 - 21:24:13 EST


On 2017/9/26 3:40, Steven Rostedt wrote:
On Mon, 11 Sep 2017 14:51:49 +0800
Zhou Chengming<zhouchengming1@xxxxxxxxxx> 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<zhouchengming1@xxxxxxxxxx>

.