Btrfs: don't allow a subvol to be deleted if it is the default subovl
authorJosef Bacik <jbacik@fusionio.com>
Mon, 12 Aug 2013 19:36:44 +0000 (15:36 -0400)
committerChris Mason <chris.mason@fusionio.com>
Sun, 1 Sep 2013 12:05:02 +0000 (08:05 -0400)
Eric pointed out that btrfs will happily allow you to delete the default subvol.
This is a problem obviously since the next time you go to mount the file system
it will freak out because it can't find the root.  Fix this by adding a check to
see if our default subvol points to the subvol we are trying to delete, and if
it does not allowing it to happen.  Thanks,

Signed-off-by: Josef Bacik <jbacik@fusionio.com>
Signed-off-by: Chris Mason <chris.mason@fusionio.com>
fs/btrfs/ioctl.c

index 022d8364e0726a169088f6a8e8b4b4326c93b017..317a984fe3c9bca6b9da6c9d17338864e12d82d2 100644 (file)
@@ -1726,13 +1726,28 @@ out:
 static noinline int may_destroy_subvol(struct btrfs_root *root)
 {
        struct btrfs_path *path;
+       struct btrfs_dir_item *di;
        struct btrfs_key key;
+       u64 dir_id;
        int ret;
 
        path = btrfs_alloc_path();
        if (!path)
                return -ENOMEM;
 
+       /* Make sure this root isn't set as the default subvol */
+       dir_id = btrfs_super_root_dir(root->fs_info->super_copy);
+       di = btrfs_lookup_dir_item(NULL, root->fs_info->tree_root, path,
+                                  dir_id, "default", 7, 0);
+       if (di && !IS_ERR(di)) {
+               btrfs_dir_item_key_to_cpu(path->nodes[0], di, &key);
+               if (key.objectid == root->root_key.objectid) {
+                       ret = -ENOTEMPTY;
+                       goto out;
+               }
+               btrfs_release_path(path);
+       }
+
        key.objectid = root->root_key.objectid;
        key.type = BTRFS_ROOT_REF_KEY;
        key.offset = (u64)-1;