Dealing with metaclass conflict with SQL Alchemy declarative base

Question

I have a class X which derives from a class with its own metaclass Meta. I want to also derive X from the declarative base in SQL Alchemy. But I can't do the simple

def class MyBase(metaclass = Meta):
    #...

def class X(declarative_base(), MyBase):
    #...

since I would get metaclass conflict error: 'the metaclass of a derived class must be a (non-strict) subclass of the metaclasses of all its bases'. I understand that I need to create a new metaclass that would derive from both Meta and from whatever metaclass the declarative base uses (DeclarativeMeta I think?). So is it enough to write:

def class NewMeta(Meta, DeclarativeMeta): pass
def class MyBase(metaclass = NewMeta):
    #...
def class X(declarative_base(), MyBase):
    #...

I tried this, and it seems to work; but I'm afraid I may have introduced some problem with this code.

I read the manual, but it's a bit too cryptic for me. What's

EDIT:

The code used for my classes is as follows:

class IterRegistry(type):
    def __new__(cls, name, bases, attr):
        attr['_registry'] = {}
        attr['_frozen'] = False
        print(name, bases)
        print(type(cls))
        return type.__new__(cls, name, bases, attr)
    def __iter__(cls):
        return iter(cls._registry.values())

class SQLEnumMeta(IterRegistry, DeclarativeMeta): pass  

class EnumType(metaclass = IterRegistry):
    def __init__(self, token):
        if hasattr(self, 'token'):
            return
        self.token = token
        self.id = len(type(self)._registry)
        type(self)._registry[token] = self

    def __new__(cls, token):
        if token in cls._registry:
            return cls._registry[token]
        else:
            if cls._frozen:
                raise TypeError('No more instances allowed')
            else:
                return object.__new__(cls)

    @classmethod
    def freeze(cls):
        cls._frozen = True

    def __repr__(self):
        return self.token

    @classmethod
    def instance(cls, token):
        return cls._registry[token]

class C1(Base, EnumType, metaclass = SQLEnumMeta):
    __tablename__ = 'c1'
    #...

Answer 1

Edit: Now having looked at IterRegistry and DeclarativeMeta, I think you're code is okay.

IterRegistry defines __new__ and __iter__, while DeclarativeMeta defines __init__ and __setattr__. Since there is no overlap, there's no direct need to call super. Nevertheless, it would good to do so, to future-proof your code.

---

Do you have control over the definition of Meta? Can you show us its definition? I don't think we can say it works or does not work unless we see the definition of Meta.

For example, there is a potential problem if your Meta does not call super(Meta,cls).__init__(classname, bases, dict_)

If you run this code

class DeclarativeMeta(type):
    def __init__(cls, classname, bases, dict_):
        print('DeclarativeMeta')
        # if '_decl_class_registry' in cls.__dict__:
        #     return type.__init__(cls, classname, bases, dict_)       
        # _as_declarative(cls, classname, dict_)
        return type.__init__(cls, classname, bases, dict_)

class Meta(type):
    def __init__(cls, classname, bases, dict_):
        print('Meta')
        return type.__init__(cls, classname, bases, dict_)

class NewMeta(Meta,DeclarativeMeta): pass

class MyBase(object):
    __metaclass__ = NewMeta
    pass

Then only the string 'Meta' gets printed. In other words, only Meta.__init__ gets run. DeclarativeMeta.__init__ gets skipped.

On the other hand, if you define

class Meta(type):
    def __init__(cls, classname, bases, dict_):
        print('Meta')
        return super(Meta,cls).__init__(classname, bases, dict_)

Then both Meta.__init__ and DeclarativeMeta.__init__ gets run.