euler inclusion
category: bugfix
bugzilla: 2535
CVE: N/A

-------------------------------------------------

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)
							  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: NZhou Chengming <zhouchengming1@huawei.com>
Signed-off-by: NKefeng Wang <wangkefeng.wang@huawei.com>
Signed-off-by: NHui Wang <john.wanghui@huawei.com>
Signed-off-by: NZhang Xiaoxu <zhangxiaoxu5@huawei.com>

Conflicts: kernel/sched/rt.c
Signed-off-by: NXuefeng Wang <wxf.wang@hisilicon.com>
Reviewed-by: NXie XiuQi <xiexiuqi@huawei.com>
Signed-off-by: NYang Yingliang <yangyingliang@huawei.com>