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>

1 files changed, 3 insertions, 2 deletions

diff --git a/fs/fs-writeback.c b/fs/fs-writeback.c
index 6bca6f8176f0..7eba7326b9b9 100644
--- 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);
 }