root = BTRFS_I(dir)->root;
+ /*
+ * 5 items for unlink inode
+ * 1 for orphan
+ */
+ ret = btrfs_reserve_metadata_space(root, 6);
+ if (ret)
+ return ret;
+
trans = btrfs_start_transaction(root, 1);
+ if (IS_ERR(trans)) {
+ btrfs_unreserve_metadata_space(root, 6);
+ return PTR_ERR(trans);
+ }
btrfs_set_trans_block_group(trans, dir);
nr = trans->blocks_used;
btrfs_end_transaction_throttle(trans, root);
+ btrfs_unreserve_metadata_space(root, 6);
btrfs_btree_balance_dirty(root, nr);
return ret;
}
inode->i_ino == BTRFS_FIRST_FREE_OBJECTID)
return -ENOTEMPTY;
+ ret = btrfs_reserve_metadata_space(root, 5);
+ if (ret)
+ return ret;
+
trans = btrfs_start_transaction(root, 1);
+ if (IS_ERR(trans)) {
+ btrfs_unreserve_metadata_space(root, 5);
+ return PTR_ERR(trans);
+ }
+
btrfs_set_trans_block_group(trans, dir);
if (unlikely(inode->i_ino == BTRFS_EMPTY_SUBVOL_DIR_OBJECTID)) {
out:
nr = trans->blocks_used;
ret = btrfs_end_transaction_throttle(trans, root);
+ btrfs_unreserve_metadata_space(root, 5);
btrfs_btree_balance_dirty(root, nr);
if (ret && !err)
return -ENOTEMPTY;
/*
- * 2 items for dir items
- * 1 item for orphan entry
- * 1 item for ref
+ * We want to reserve the absolute worst case amount of items. So if
+ * both inodes are subvols and we need to unlink them then that would
+ * require 4 item modifications, but if they are both normal inodes it
+ * would require 5 item modifications, so we'll assume their normal
+ * inodes. So 5 * 2 is 10, plus 1 for the new link, so 11 total items
+ * should cover the worst case number of items we'll modify.
*/
- ret = btrfs_reserve_metadata_space(root, 4);
+ ret = btrfs_reserve_metadata_space(root, 11);
if (ret)
return ret;
if (old_inode->i_ino == BTRFS_FIRST_FREE_OBJECTID)
up_read(&root->fs_info->subvol_sem);
- btrfs_unreserve_metadata_space(root, 4);
+ btrfs_unreserve_metadata_space(root, 11);
return ret;
}