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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>
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Modificáronse 1 ficheiros con 3 adicións e 2 borrados
  1. 3 2
      fs/fs-writeback.c

+ 3 - 2
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);
 }