Oleg noticed that its possible to falsely observe p->on_cpu == 0 such
that we'll prematurely continue with the wakeup and effectively run p on
two CPUs at the same time.

Even though the overlap is very limited; the task is in the middle of
being scheduled out; it could still result in corruption of the
scheduler data structures.

        CPU0                            CPU1

        set_current_state(...)

        <preempt_schedule>
          context_switch(X, Y)
            prepare_lock_switch(Y)
              Y->on_cpu = 1;
            finish_lock_switch(X)
              store_release(X->on_cpu, 0);

                                        try_to_wake_up(X)
                                          LOCK(p->pi_lock);

                                          t = X->on_cpu; // 0

          context_switch(Y, X)
            prepare_lock_switch(X)
              X->on_cpu = 1;
            finish_lock_switch(Y)
              store_release(Y->on_cpu, 0);
        </preempt_schedule>

        schedule();
          deactivate_task(X);
          X->on_rq = 0;

                                          if (X->on_rq) // false

                                          if (t) while (X->on_cpu)
                                            cpu_relax();

          context_switch(X, ..)
            finish_lock_switch(X)
              store_release(X->on_cpu, 0);

Avoid the load of X->on_cpu being hoisted over the X->on_rq load.
Reported-by: NOleg Nesterov <oleg@redhat.com>
Signed-off-by: NPeter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Signed-off-by: NIngo Molnar <mingo@kernel.org>