On Tue 21-12-21 11:21:35, Yu Kuai wrote:
During code review, we found that if bfqq is not busy in
bfq_bfqq_move(), bfq_pos_tree_add_move() won't be called for the bfqq,
thus bfqq->pos_root still points to the old bfqg. However, the ref
that bfqq hold for the old bfqg will be released, so it's possible
that the old bfqg can be freed. This is problematic because the freed
bfqg can still be accessed by bfqq->pos_root.
Fix the problem by calling bfq_pos_tree_add_move() for idle bfqq
as well.
Fixes: e21b7a0b9887 ("block, bfq: add full hierarchical scheduling and cgroups support")
Signed-off-by: Yu Kuai <yukuai3@xxxxxxxxxx>
I'm just wondering, how can it happen that !bfq_bfqq_busy() queue is in
pos_tree? Because bfq_remove_request() takes care to remove bfqq from the
pos_tree...
Honza
---
block/bfq-cgroup.c | 14 +++++++++-----
1 file changed, 9 insertions(+), 5 deletions(-)
diff --git a/block/bfq-cgroup.c b/block/bfq-cgroup.c
index 8e8cf6b3d946..822dd28ecf53 100644
--- a/block/bfq-cgroup.c
+++ b/block/bfq-cgroup.c
@@ -677,7 +677,6 @@ void bfq_bfqq_move(struct bfq_data *bfqd, struct bfq_queue *bfqq,
bfq_deactivate_bfqq(bfqd, bfqq, false, false);
else if (entity->on_st_or_in_serv)
bfq_put_idle_entity(bfq_entity_service_tree(entity), entity);
- bfqg_and_blkg_put(old_parent);
if (entity->parent &&
entity->parent->last_bfqq_created == bfqq)
@@ -690,11 +689,16 @@ void bfq_bfqq_move(struct bfq_data *bfqd, struct bfq_queue *bfqq,
/* pin down bfqg and its associated blkg */
bfqg_and_blkg_get(bfqg);
- if (bfq_bfqq_busy(bfqq)) {
- if (unlikely(!bfqd->nonrot_with_queueing))
- bfq_pos_tree_add_move(bfqd, bfqq);
+ /*
+ * Don't leave the pos_root to old bfqg, since the ref to old bfqg will
+ * be released and the bfqg might be freed.
+ */
+ if (unlikely(!bfqd->nonrot_with_queueing))
+ bfq_pos_tree_add_move(bfqd, bfqq);
+ bfqg_and_blkg_put(old_parent);
+
+ if (bfq_bfqq_busy(bfqq))
bfq_activate_bfqq(bfqd, bfqq);
- }
if (!bfqd->in_service_queue && !bfqd->rq_in_driver)
bfq_schedule_dispatch(bfqd);
--
2.31.1