Consider a new I/O request that arrives for a bfq_queue bfqq. If, when
this happens, the only active bfq_queues are bfqq and either its waker
bfq_queue or one of its woken bfq_queues, then there is no point in
queueing this new I/O request in bfqq for service. In fact, the
in-service queue and bfqq agree on serving this new I/O request as
soon as possible. So this commit puts this new I/O request directly
into the dispatch list.
Tested-by: Jan Kara <jack@xxxxxxx>
Signed-off-by: Paolo Valente <paolo.valente@xxxxxxxxxx>
---
block/bfq-iosched.c | 17 ++++++++++++++++-
1 file changed, 16 insertions(+), 1 deletion(-)
diff --git a/block/bfq-iosched.c b/block/bfq-iosched.c
index a83149407336..e5b83910fbe0 100644
--- a/block/bfq-iosched.c
+++ b/block/bfq-iosched.c
@@ -5640,7 +5640,22 @@ static void bfq_insert_request(struct blk_mq_hw_ctx *hctx, struct request *rq,
spin_lock_irq(&bfqd->lock);
bfqq = bfq_init_rq(rq);
- if (!bfqq || at_head || blk_rq_is_passthrough(rq)) {
+
+ /*
+ * Additional case for putting rq directly into the dispatch
+ * queue: the only active bfq_queues are bfqq and either its
+ * waker bfq_queue or one of its woken bfq_queues. In this
+ * case, there is no point in queueing rq in bfqq for
+ * service. In fact, the in-service queue and bfqq agree on
+ * serving this new I/O request as soon as possible.
+ */
+ if (!bfqq ||
+ (bfqq != bfqd->in_service_queue &&
+ bfqd->in_service_queue != NULL &&
+ bfq_tot_busy_queues(bfqd) == 1 + bfq_bfqq_busy(bfqq) &&
+ (bfqq->waker_bfqq == bfqd->in_service_queue ||
+ bfqd->in_service_queue->waker_bfqq == bfqq)) ||
+ at_head || blk_rq_is_passthrough(rq)) {
if (at_head)
list_add(&rq->queuelist, &bfqd->dispatch);
else
--
2.20.1
