[PATCH] sched/fair: Simplify the scene where both local and busiest are group_fully_busy

[zgpeng]

When both local and busiest group are group_fully_busy type, because
the avg_load of the group of type group_fully_busy is not calculated, the
avg_load value is equal to 0. In this case, load balancing will not actually 
done, because after a series of calculations in the calculate_imbalance, it 
will be considered that load balance is not required. Therefore,it is not
necessary to enter calculate_imbalance to do some useless work.

Signed-off-by: zgpeng <zgpeng@xxxxxxxxxxx>
Reviewed-by: Samuel Liao <samuelliao@xxxxxxxxxxx>
---
 kernel/sched/fair.c | 12 ++++++++++++
 1 file changed, 12 insertions(+)

diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
index 9f75303..cc1d6c4 100644
--- a/kernel/sched/fair.c
+++ b/kernel/sched/fair.c
@@ -9634,6 +9634,18 @@ static struct sched_group *find_busiest_group(struct lb_env *env)
 			 * busiest doesn't have any tasks waiting to run
 			 */
 			goto out_balanced;
+
+		if (local->group_type == group_fully_busy)
+			/*
+			 * If local group is group_fully_busy, the code goes here,
+			 * the type of busiest group must also be group_fully_busy.
+			 * Because the avg_load of the group_fully_busy type is not
+			 * calculated at present, it is actually equal to 0. In this
+			 * scenario, load balance is not performed. therefore, it can
+			 * be returned directly here, and there is no need to do some
+			 * useless work in calculate_imbalance.
+			 */
+			goto out_balanced;
 	}
 
 force_balance:
-- 
2.9.5