audit: fix a memory leak bug

In audit_rule_change(), audit_data_to_entry() is firstly invoked to
translate the payload data to the kernel's rule representation. In
audit_data_to_entry(), depending on the audit field type, an audit tree may
be created in audit_make_tree(), which eventually invokes kmalloc() to
allocate the tree.  Since this tree is a temporary tree, it will be then
freed in the following execution, e.g., audit_add_rule() if the message
type is AUDIT_ADD_RULE or audit_del_rule() if the message type is
AUDIT_DEL_RULE. However, if the message type is neither AUDIT_ADD_RULE nor
AUDIT_DEL_RULE, i.e., the default case of the switch statement, this
temporary tree is not freed.

To fix this issue, only allocate the tree when the type is AUDIT_ADD_RULE
or AUDIT_DEL_RULE.
Signed-off-by: Wenwen Wang <wang6495@umn.edu>
Reviewed-by: Richard Guy Briggs <rgb@redhat.com>
Signed-off-by: Paul Moore <paul@paul-moore.com>

kernel/auditfilter.c

@@ -1114,22 +1114,24 @@ int audit_rule_change(int type, int seq, void *data, size_t datasz)
 	int err = 0;
 	struct audit_entry *entry;
 
-	entry = audit_data_to_entry(data, datasz);
-	if (IS_ERR(entry))
-		return PTR_ERR(entry);
-
 	switch (type) {
 	case AUDIT_ADD_RULE:
+		entry = audit_data_to_entry(data, datasz);
+		if (IS_ERR(entry))
+			return PTR_ERR(entry);
 		err = audit_add_rule(entry);
 		audit_log_rule_change("add_rule", &entry->rule, !err);
 		break;
 	case AUDIT_DEL_RULE:
+		entry = audit_data_to_entry(data, datasz);
+		if (IS_ERR(entry))
+			return PTR_ERR(entry);
 		err = audit_del_rule(entry);
 		audit_log_rule_change("remove_rule", &entry->rule, !err);
 		break;
 	default:
-		err = -EINVAL;
 		WARN_ON(1);
+		return -EINVAL;
 	}
 
 	if (err || type == AUDIT_DEL_RULE) {