sched, lockdep: Annotate ->pi_lock recursion

There's a valid ->pi_lock recursion issue where the actual PI code
tries to wake up the stop task. Make lockdep aware so it doesn't
complain about this.

Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Reviewed-by: Valentin Schneider <valentin.schneider@arm.com>
Reviewed-by: Daniel Bristot de Oliveira <bristot@redhat.com>
Link: https://lkml.kernel.org/r/20201023102347.406912197@infradead.org

kernel/sched/core.c

@@ -2658,6 +2658,7 @@ int select_task_rq(struct task_struct *p, int cpu, int sd_flags, int wake_flags)
 
 void sched_set_stop_task(int cpu, struct task_struct *stop)
 {
+	static struct lock_class_key stop_pi_lock;
 	struct sched_param param = { .sched_priority = MAX_RT_PRIO - 1 };
 	struct task_struct *old_stop = cpu_rq(cpu)->stop;
 
@@ -2673,6 +2674,20 @@ void sched_set_stop_task(int cpu, struct task_struct *stop)
 		sched_setscheduler_nocheck(stop, SCHED_FIFO, &param);
 
 		stop->sched_class = &stop_sched_class;
+
+		/*
+		 * The PI code calls rt_mutex_setprio() with ->pi_lock held to
+		 * adjust the effective priority of a task. As a result,
+		 * rt_mutex_setprio() can trigger (RT) balancing operations,
+		 * which can then trigger wakeups of the stop thread to push
+		 * around the current task.
+		 *
+		 * The stop task itself will never be part of the PI-chain, it
+		 * never blocks, therefore that ->pi_lock recursion is safe.
+		 * Tell lockdep about this by placing the stop->pi_lock in its
+		 * own class.
+		 */
+		lockdep_set_class(&stop->pi_lock, &stop_pi_lock);
 	}
 
 	cpu_rq(cpu)->stop = stop;