[1/3] fuse: Kill fasync only if interrupt is queued in queue_interrupt()

Submitted by Kirill Tkhai on April 2, 2019, 9:16 a.m.

Details

Message ID 155419659215.4212.15152211256672184767.stgit@localhost.localdomain
State New
Series "fuse: Backport of ms locking patches part 1"
Headers show

Commit Message

Kirill Tkhai April 2, 2019, 9:16 a.m.
ms commit 8da6e9183275

We should sent signal only in case of interrupt is really queued.  Not a
real problem, but this makes the code clearer and intuitive.

Signed-off-by: Kirill Tkhai <ktkhai@virtuozzo.com>
Signed-off-by: Miklos Szeredi <mszeredi@redhat.com>
Signed-off-by: Kirill Tkhai <ktkhai@virtuozzo.com>
---
 fs/fuse/dev.c |    2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

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diff --git a/fs/fuse/dev.c b/fs/fuse/dev.c
index dc21886ee55d..eb5c62e3170f 100644
--- a/fs/fuse/dev.c
+++ b/fs/fuse/dev.c
@@ -479,9 +479,9 @@  static void queue_interrupt(struct fuse_iqueue *fiq, struct fuse_req *req)
 	if (list_empty(&req->intr_entry)) {
 		list_add_tail(&req->intr_entry, &fiq->interrupts);
 		wake_up_locked(&fiq->waitq);
+		kill_fasync(&fiq->fasync, SIGIO, POLL_IN);
 	}
 	spin_unlock(&fiq->waitq.lock);
-	kill_fasync(&fiq->fasync, SIGIO, POLL_IN);
 }
 
 static void request_wait_answer(struct fuse_conn *fc, struct fuse_req *req)