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show proper interrogate name instead of 'type'

David Rose %!s(int64=19) %!d(string=hai) anos
pai
achega
8eb5baa427
Modificáronse 1 ficheiros con 1 adicións e 1 borrados
  1. 1 1
      dtool/src/interrogatedb/py_panda.cxx

+ 1 - 1
dtool/src/interrogatedb/py_panda.cxx

@@ -82,7 +82,7 @@ DTOOL_Call_GetPointerThisClass(PyObject *self, Dtool_PyTypedObject *classdef,
         str << function_name << "() argument " << param << " must be ";
         str << function_name << "() argument " << param << " must be ";
 
 
 
 
-        PyObject *fname = PyObject_GetAttrString((PyObject *)classdef->_PyType.ob_type, "__name__");
+        PyObject *fname = PyObject_GetAttrString((PyObject *)classdef, "__name__");
         if (fname != (PyObject *)NULL) {
         if (fname != (PyObject *)NULL) {
           str << PyString_AsString(fname);
           str << PyString_AsString(fname);
           Py_DECREF(fname);
           Py_DECREF(fname);