Btrfs: detect corruption when non-root leaf has zero item
Right now we treat leaf which has zero item as a valid one because we could have an empty tree, that is, a root that is also a leaf without any item, however, in the same case but when the leaf is not a root, we can end up with hitting the BUG_ON(1) in btrfs_extend_item() called by setup_inline_extent_backref(). This makes us check the situation as a corruption if leaf is not its own root. Signed-off-by: Liu Bo <bo.li.liu@oracle.com> Reviewed-by: David Sterba <dsterba@suse.com> Signed-off-by: David Sterba <dsterba@suse.com> Signed-off-by: Chris Mason <clm@fb.com>
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@ -560,8 +560,29 @@ static noinline int check_leaf(struct btrfs_root *root,
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u32 nritems = btrfs_header_nritems(leaf);
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int slot;
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if (nritems == 0)
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if (nritems == 0) {
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struct btrfs_root *check_root;
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key.objectid = btrfs_header_owner(leaf);
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key.type = BTRFS_ROOT_ITEM_KEY;
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key.offset = (u64)-1;
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check_root = btrfs_get_fs_root(root->fs_info, &key, false);
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/*
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* The only reason we also check NULL here is that during
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* open_ctree() some roots has not yet been set up.
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*/
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if (!IS_ERR_OR_NULL(check_root)) {
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/* if leaf is the root, then it's fine */
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if (leaf->start !=
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btrfs_root_bytenr(&check_root->root_item)) {
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CORRUPT("non-root leaf's nritems is 0",
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leaf, root, 0);
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return -EIO;
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}
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}
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return 0;
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}
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/* Check the 0 item */
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if (btrfs_item_offset_nr(leaf, 0) + btrfs_item_size_nr(leaf, 0) !=
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