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

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>

diff --git a/fs/btrfs/reada.c b/fs/btrfs/reada.c
index ee6f60547a8d9eeb67d593bc76ce3f1c197524fa..dd4f9c2b7107474e4513a79b403562c9fce4c480 100644 (file)
--- a/fs/btrfs/reada.c
+++ b/fs/btrfs/reada.c
@@ -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)