python 中是否有一种神奇的方法可以计算一个类?

Question

我有一个mixin类，只有在与其他类一起使用时才应该用作接口，例如：

class Mother():
pass
class Child(Mother):
pass
class Mixin():
def __init__(self):
assert isinstance(self, Mother), 'Mixin can only be used with Mother implementations'
super().__init__()
class Implementation(Mixin, Child):
pass
Implementation()

上述方法有效，但只有在实例化Implementation时，我才能以某种方式在代码执行时评估上述断言？

这一点很重要，这样如果有人错误地实现了类，应用程序就不会运行。

(我不确定我的标题措辞是否正确(

Answer 1

实际上，即使"当Implementation被实例化时"，它也不会工作 - 它会Child通过类找到与Mother类的关系(Implementation继承Child--->Child继承Mother(，
因此isinstance(self, Mother)由于继承链(被认为是mro(方法解析顺序((Mother类Implementation视为派生__init_subclass__钩子
：

class Mother():
pass    
class Child(Mother):
pass    
class Mixin():
def __init_subclass__(cls, **kwargs):
assert isinstance(cls, Mother), 'Mixin can only be used with Mother'
super().__init_subclass__(**kwargs)    
class Implementation(Mixin, Child):
pass    
Implementation()

抛出：

Traceback (most recent call last):
File ..., in __init_subclass__
assert isinstance(cls, Mother), 'Mixin can only be used with Mother'
AssertionError: Mixin can only be used with Mother

---

但是，如果您需要允许将Mixin应用于类及其子类Mother- 请改用issubclass调用：

class Mixin():
def __init_subclass__(cls, **kwargs):
assert issubclass(cls, Mother), 'Mixin can only be used with Mother and its subclasses'
super().__init_subclass__(**kwargs)

钩子将应用于类声明阶段(在潜在实例化之前(

Answer 2

你也可以使用元类，它很强大，可以帮助你理解python类。

class Mother():
pass    
class Child(Mother):
pass    

class Meta(type):
def __new__(meta_cls, name, bases, dct):
if name != "Mixin" and all([not issubclass(b, Mother) for b in bases]):
raise Exception('Mixin can only be used with Mother')
cls = super().__new__(meta_cls, name, bases, dct)
return cls
class Mixin(metaclass=Meta):
pass  
class Implementation(Mixin, Child):
pass