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btrfs: avoid unnecessary path allocation at fixup_inode_link_count()
There's no need to allocate a path as our single caller already has a path that we can use. So pass the caller's path and use it. Signed-off-by: Filipe Manana <fdmanana@suse.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com>
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parent
2ac7094662
commit
29d9c5e037
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@ -1728,19 +1728,15 @@ static int count_inode_refs(struct btrfs_inode *inode, struct btrfs_path *path)
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* will free the inode.
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*/
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static noinline int fixup_inode_link_count(struct walk_control *wc,
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struct btrfs_path *path,
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struct btrfs_inode *inode)
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{
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struct btrfs_trans_handle *trans = wc->trans;
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struct btrfs_root *root = inode->root;
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struct btrfs_path *path;
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int ret;
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u64 nlink = 0;
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const u64 ino = btrfs_ino(inode);
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path = btrfs_alloc_path();
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if (!path)
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return -ENOMEM;
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ret = count_inode_refs(inode, path);
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if (ret < 0)
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goto out;
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@ -1776,7 +1772,7 @@ static noinline int fixup_inode_link_count(struct walk_control *wc,
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}
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out:
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btrfs_free_path(path);
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btrfs_release_path(path);
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return ret;
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}
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@ -1821,7 +1817,7 @@ static noinline int fixup_inode_link_counts(struct walk_control *wc,
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break;
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}
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ret = fixup_inode_link_count(wc, inode);
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ret = fixup_inode_link_count(wc, path, inode);
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iput(&inode->vfs_inode);
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if (ret)
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break;
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