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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>
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@ -752,21 +752,19 @@ static int reada_start_machine_dev(struct btrfs_device *dev)
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static void reada_start_machine_worker(struct btrfs_work *work)
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{
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struct reada_machine_work *rmw;
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struct btrfs_fs_info *fs_info;
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int old_ioprio;
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rmw = container_of(work, struct reada_machine_work, work);
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fs_info = rmw->fs_info;
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kfree(rmw);
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old_ioprio = IOPRIO_PRIO_VALUE(task_nice_ioclass(current),
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task_nice_ioprio(current));
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set_task_ioprio(current, BTRFS_IOPRIO_READA);
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__reada_start_machine(fs_info);
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__reada_start_machine(rmw->fs_info);
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set_task_ioprio(current, old_ioprio);
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atomic_dec(&fs_info->reada_works_cnt);
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atomic_dec(&rmw->fs_info->reada_works_cnt);
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kfree(rmw);
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}
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static void __reada_start_machine(struct btrfs_fs_info *fs_info)
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