sched/deadline: Fix confusing comments about selection of top pi-waiter

This comment in the code is incomplete, and I believe it begs a definition of
dl_boosted to make sense of the condition that follows. Rewrite the comment and
also rearrange the condition that follows to reflect the first condition "we
have a top pi-waiter which is a SCHED_DEADLINE task" in that order. Also fix a
typo that follows.

Signed-off-by: Joel Fernandes <joelaf@google.com>
Reviewed-by: Daniel Bristot de Oliveira <bristot@redhat.com>
Acked-by: Juri Lelli <juri.lelli@arm.com>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Link: http://lkml.kernel.org/r/20170713022429.10307-1-joelaf@google.com
Signed-off-by: Ingo Molnar <mingo@kernel.org>
diff --git a/kernel/sched/deadline.c b/kernel/sched/deadline.c
index a84299f..755bd3f 100644
--- a/kernel/sched/deadline.c
+++ b/kernel/sched/deadline.c
@@ -1392,17 +1392,19 @@ static void enqueue_task_dl(struct rq *rq, struct task_struct *p, int flags)
 	struct sched_dl_entity *pi_se = &p->dl;
 
 	/*
-	 * Use the scheduling parameters of the top pi-waiter
-	 * task if we have one and its (absolute) deadline is
-	 * smaller than our one... OTW we keep our runtime and
-	 * deadline.
+	 * Use the scheduling parameters of the top pi-waiter task if:
+	 * - we have a top pi-waiter which is a SCHED_DEADLINE task AND
+	 * - our dl_boosted is set (i.e. the pi-waiter's (absolute) deadline is
+	 *   smaller than our deadline OR we are a !SCHED_DEADLINE task getting
+	 *   boosted due to a SCHED_DEADLINE pi-waiter).
+	 * Otherwise we keep our runtime and deadline.
 	 */
-	if (pi_task && p->dl.dl_boosted && dl_prio(pi_task->normal_prio)) {
+	if (pi_task && dl_prio(pi_task->normal_prio) && p->dl.dl_boosted) {
 		pi_se = &pi_task->dl;
 	} else if (!dl_prio(p->normal_prio)) {
 		/*
 		 * Special case in which we have a !SCHED_DEADLINE task
-		 * that is going to be deboosted, but exceedes its
+		 * that is going to be deboosted, but exceeds its
 		 * runtime while doing so. No point in replenishing
 		 * it, as it's going to return back to its original
 		 * scheduling class after this.