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>
This commit is contained in:
Nick Piggin 2009-09-15 21:34:51 +02:00 committed by Jens Axboe
parent 77fad5e625
commit 77b9d059cb
1 changed files with 3 additions and 2 deletions

View File

@ -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);
}