python - 我怎样才能强制子类有 __slots__？

Question

我有一节课 __slots__ :

class A:
    __slots__ = ('foo',)

如果我创建一个子类而不指定__slots__，子类将有一个__dict__:

class B(A):
    pass

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

有什么方法可以防止 B 有一个 __dict__ 而不必设置 __slots__ = ()？

Answer 1

@AKX的答案几乎是正确的。我认为 __prepare__ 和元类确实是很容易解决这个问题的方法。

简单回顾一下:

• 如果类的命名空间在类主体执行后包含 __slots__ 键，则该类将使用 __slots__ 而不是 __dict__。
• 可以使用 __prepare__ 将名称注入(inject)到类的命名空间之前。

因此，如果我们简单地从 __prepare__ 返回包含键 '__slots__' 的字典，那么类将(如果 '__slots__' 键在类主体的评估期间不会再次删除)使用 __slots__ 而不是 __dict__。因为 __prepare__ 只是提供了初始命名空间，所以可以轻松覆盖 __slots__ 或在类主体中再次删除它们。

因此默认情况下提供 __slots__ 的元类看起来像这样:

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

因此，每个具有此元类的类和子类(默认情况下)在其命名空间中都有一个空的 __slots__，从而创建一个“带槽的类”(除了 __slots__被故意删除)。

只是为了说明这是如何工作的:

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__",

通过了 AKZ 回答中提到的测试:

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)

并验证它是否按预期工作:

# 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

正如评论(作者 Aran-Fey)所指出的，del __slots__ 和将 __dict__ 添加到 __slots__ 之间是有区别的:

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