writeback: Fix bdi use after free in wb_work_complete()

By the time bdi_work_on_stack gets evaluated again in bdi_work_free, it
can already have been deallocated and used for something else in the
!on stack case, giving a false positive in this test and causing
corruption.

Signed-off-by: Nick Piggin <npiggin@suse.de>
Signed-off-by: Jens Axboe <jens.axboe@oracle.com>

diff --git a/fs/fs-writeback.c b/fs/fs-writeback.c
index 6bca6f8176f0d8f105be5379b31837bf85f27da1..7eba7326b9b95568913ef0b774a44b20bd4b8e5f 100644 (file)
--- a/fs/fs-writeback.c
+++ b/fs/fs-writeback.c
@@ -113,6 +113,7 @@ static void bdi_work_free(struct rcu_head *head)
 static void wb_work_complete(struct bdi_work *work)
 {
        const enum writeback_sync_modes sync_mode = work->args.sync_mode;
+       int onstack = bdi_work_on_stack(work);
 
        /*
         * For allocated work, we can clear the done/seen bit right here.
@@ -120,9 +121,9 @@ static void wb_work_complete(struct bdi_work *work)
         * to after the RCU grace period, since the stack could be invalidated
         * as soon as bdi_work_clear() has done the wakeup.
         */
-       if (!bdi_work_on_stack(work))
+       if (!onstack)
                bdi_work_clear(work);
-       if (sync_mode == WB_SYNC_NONE || bdi_work_on_stack(work))
+       if (sync_mode == WB_SYNC_NONE || onstack)
                call_rcu(&work->rcu_head, bdi_work_free);
 }