How can I force subclasses to have __slots__?

Question

I have a class with __slots__:

class A:
    __slots__ = ('foo',)

If I create a subclass without specifying __slots__, the subclass will have a __dict__:

class B(A):
    pass

print('__dict__' in dir(B))  # True

Is there any way to prevent B from having a __dict__ without having to set __slots__ = ()?

Answer 1

The answer of @AKX is almost correct. I think __prepare__ and a metaclass is indeed the way this can be solved quite easily.

Just to recap:

• If the namespace of the class contains a __slots__ key after the class body is executed then the class will use __slots__ instead of __dict__.
• One can inject names into the namespace of the class before the class body is executed by using __prepare__.

So if we simply return a dictionary containing the key '__slots__' from __prepare__ then the class will (if the '__slots__' key isn't removed again during the evaluation of the class body) use __slots__ instead of __dict__. Because __prepare__ just provides the initial namespace one can easily override the __slots__ or remove them again in the class body.

So a metaclass that provides __slots__ by default would look like this:

class ForceSlots(type):
    @classmethod
    def __prepare__(metaclass, name, bases, **kwds):
        # calling super is not strictly necessary because
        #  type.__prepare() simply returns an empty dict.
        # But if you plan to use metaclass-mixins then this is essential!
        super_prepared = super().__prepare__(metaclass, name, bases, **kwds)
        super_prepared['__slots__'] = ()
        return super_prepared

So every class and subclass with this metaclass will (by default) have an empty __slots__ in their namespace and thus create a "class with slots" (except the __slots__ are removed on purpose).

Just to illustrate how this would work:

class A(metaclass=ForceSlots):
    __slots__ = "a",

class B(A):  # no __dict__ even if slots are not defined explicitly
    pass

class C(A):  # no __dict__, but provides additional __slots__
    __slots__ = "c",

class D(A):  # creates normal __dict__-based class because __slots__ was removed
    del __slots__

class E(A):  # has a __dict__ because we added it to __slots__
    __slots__ = "__dict__",

Which passes the tests mentioned in AKZs answer:

assert "__dict__" not in dir(A)
assert "__dict__" not in dir(B)
assert "__dict__" not in dir(C)
assert "__dict__" in dir(D)
assert "__dict__" in dir(E)

And to verify that it works as expected:

# A has slots from A: a
a = A()
a.a = 1
a.b = 1  # AttributeError: 'A' object has no attribute 'b'

# B has slots from A: a
b = B()  
b.a = 1
b.b = 1  # AttributeError: 'B' object has no attribute 'b'

# C has the slots from A and C: a and c
c = C()
c.a = 1
c.b = 1  # AttributeError: 'C' object has no attribute 'b'
c.c = 1

# D has a dict and allows any attribute name
d = D()  
d.a = 1
d.b = 1
d.c = 1

# E has a dict and allows any attribute name
e = E()  
e.a = 1
e.b = 1
e.c = 1

As pointed out in a comment (by Aran-Fey) there is a difference between del __slots__ and adding __dict__ to the __slots__:

There's a minor difference between the two options: del __slots__ will give your class not only a __dict__, but also a __weakref__ slot.