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btrfs: Remove delayed_iput member from btrfs_delalloc_work
When allocating a delalloc work we are always setting the delayed_iput to 0. So remove the delay_iput member of btrfs_delalloc_work, as a result also remove it as a parameter from btrfs_alloc_delalloc_work since it's not used anymore. Signed-off-by: Nikolay Borisov <nborisov@suse.com> Reviewed-by: Qu Wenruo <wqu@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
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@ -3169,14 +3169,12 @@ void btrfs_extent_item_to_extent_map(struct btrfs_inode *inode,
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/* inode.c */
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struct btrfs_delalloc_work {
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struct inode *inode;
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int delay_iput;
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struct completion completion;
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struct list_head list;
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struct btrfs_work work;
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};
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struct btrfs_delalloc_work *btrfs_alloc_delalloc_work(struct inode *inode,
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int delay_iput);
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struct btrfs_delalloc_work *btrfs_alloc_delalloc_work(struct inode *inode);
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struct extent_map *btrfs_get_extent_fiemap(struct btrfs_inode *inode,
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struct page *page, size_t pg_offset, u64 start,
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@ -10159,15 +10159,11 @@ static void btrfs_run_delalloc_work(struct btrfs_work *work)
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&BTRFS_I(inode)->runtime_flags))
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filemap_flush(inode->i_mapping);
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if (delalloc_work->delay_iput)
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btrfs_add_delayed_iput(inode);
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else
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iput(inode);
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iput(inode);
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complete(&delalloc_work->completion);
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}
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struct btrfs_delalloc_work *btrfs_alloc_delalloc_work(struct inode *inode,
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int delay_iput)
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struct btrfs_delalloc_work *btrfs_alloc_delalloc_work(struct inode *inode)
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{
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struct btrfs_delalloc_work *work;
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@ -10178,7 +10174,6 @@ struct btrfs_delalloc_work *btrfs_alloc_delalloc_work(struct inode *inode,
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init_completion(&work->completion);
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INIT_LIST_HEAD(&work->list);
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work->inode = inode;
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work->delay_iput = delay_iput;
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WARN_ON_ONCE(!inode);
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btrfs_init_work(&work->work, btrfs_flush_delalloc_helper,
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btrfs_run_delalloc_work, NULL, NULL);
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@ -10218,7 +10213,7 @@ static int start_delalloc_inodes(struct btrfs_root *root, int nr)
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}
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spin_unlock(&root->delalloc_lock);
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work = btrfs_alloc_delalloc_work(inode, 0);
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work = btrfs_alloc_delalloc_work(inode);
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if (!work) {
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iput(inode);
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ret = -ENOMEM;
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