[issue15870] PyType_FromSpec should take metaclass as an argument

[Josh Haberman]

Josh Haberman <jhaberman at gmail.com> added the comment:

> You can also call (PyObject_Call*) the metaclass with (name, bases, namespace);

But won't that just call my metaclass's tp_new?  I'm trying to do this from my metaclass's tp_new, so I can customize the class creation process. Then Python code can use my metaclass to construct classes normally.

> I wouldn't recommend [setting ob_type] after PyType_Ready is called.

Why not?  What bad things will happen?  It seems to be working so far.

Setting ob_type directly actually solves another problem that I had been having with the limited API.  I want to implement tp_getattro on the metaclass, but I want to first delegate to PyType.tp_getattro to return any entry that may be present in the type's tp_dict.  With the full API I could call self->ob_type->tp_base->tp_getattro() do to the equivalent of super(), but with the limited API I can't access type->tp_getattro (and PyType_GetSlot() can't be used on non-heap types).

I find that this does what I want:

  PyTypeObject *saved_type = self->ob_type;
  self->ob_type = &PyType_Type;
  PyObject *ret = PyObject_GetAttr(self, name);
  self->ob_type = saved_type;

Previously I had tried:

   PyObject *super = PyObject_CallFunction((PyObject *)&PySuper_Type, "OO",
                                           self->ob_type, self);
   PyObject *ret = PyObject_GetAttr(super, name);
   Py_DECREF(super);

But for some reason this didn't work.

----------

_______________________________________
Python tracker <report at bugs.python.org>
<https://bugs.python.org/issue15870>
_______________________________________