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btrfs: don't prematurely free work in reada_start_machine_worker() 

[ Upstream commit e732fe95e4 ]

Currently, reada_start_machine_worker() frees the reada_machine_work and
then calls __reada_start_machine() to do readahead. This is another
potential instance of the bug in "btrfs: don't prematurely free work in
run_ordered_work()".

There _might_ already be a deadlock here: reada_start_machine_worker()
can depend on itself through stacked filesystems (__read_start_machine()
-> reada_start_machine_dev() -> reada_tree_block_flagged() ->
read_extent_buffer_pages() -> submit_one_bio() ->
btree_submit_bio_hook() -> btrfs_map_bio() -> submit_stripe_bio() ->
submit_bio() onto a loop device can trigger readahead on the lower
filesystem).

Either way, let's fix it by freeing the work at the end.

Reviewed-by: Johannes Thumshirn <jthumshirn@suse.de>
Signed-off-by: Omar Sandoval <osandov@fb.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
Signed-off-by: Sasha Levin <sashal@kernel.org>
This commit is contained in:
Omar Sandoval 2019-09-16 11:30:55 -07:00 committed by Greg Kroah-Hartman
parent dc9d0a75ce
commit 736d28c704
1 changed files with 4 additions and 6 deletions
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10
fs/btrfs/reada.c
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@ -752,21 +752,19 @@ static int reada_start_machine_dev(struct btrfs_device *dev)
static void reada_start_machine_worker(struct btrfs_work *work)
{
struct reada_machine_work *rmw;
struct btrfs_fs_info *fs_info;
int old_ioprio;
rmw = container_of(work, struct reada_machine_work, work);
fs_info = rmw->fs_info;
kfree(rmw);
old_ioprio = IOPRIO_PRIO_VALUE(task_nice_ioclass(current),
task_nice_ioprio(current));
set_task_ioprio(current, BTRFS_IOPRIO_READA);
__reada_start_machine(fs_info);
__reada_start_machine(rmw->fs_info);
set_task_ioprio(current, old_ioprio);
atomic_dec(&fs_info->reada_works_cnt);
atomic_dec(&rmw->fs_info->reada_works_cnt);
kfree(rmw);
}
static void __reada_start_machine(struct btrfs_fs_info *fs_info)