python - 如何在类中声明装饰器来装饰已经装饰的继承方法？

Question

我想创建一个装饰器，它是一个类成员，它将装饰一个继承的方法，该方法被装饰。

示例代码:

class A(object):
    __metaclass__ = ABCMeta
    def __init__(self):
        pass

    @classmethod
    def the_decorator(cls, decorated):  # <-----this is what i want, with or without self/cls as an argument
        def decorator()
            #do stuff before
            decorated()
            print "decorator was called!"
            #do stuff after
        return decorator

    @abstractmethod
    def inherited():
        raise NotImplemented


class B(A):
    def __init__(self):
        super(B,self).__init__()

    #@A.the_decorator <--- this is what I want, 
    @overrides
    #@A.the_decorator <--- or this
    def inherited():
        print "B.inherited was invoked"

和

b = B()
b.inherited()

应该输出

B.inherited was invoked

decorator was called!

<小时/>

已阅读 this guide on decorators as class members ，我仍然无法弄清楚如何使用父类(super class)中定义的装饰器来装饰继承的方法。

<小时/>

注意，这里@overrides是由overridespackage定义的。 pip install 覆盖

<小时/>

另请注意，我目前使用的是 python 2.7，但会喜欢 2.7 和 3+ 答案。

谢谢!

Answer 1

你还没有那么远!

关键是装饰器将接收一个参数，该参数是装饰函数，因此它只能是一种统计方法。而且您还忘记了应该使用 self 参数声明普通方法。

但是这段代码应该可以工作:

class A(object):
    __metaclass__ = ABCMeta
    def __init__(self):
        pass

    @staticmethod
    def the_decorator(decorated):  # <-----this is what i want, with or without self/cls as an argument
        def decorator(self):
            #do stuff before
            decorated(self)
            print "decorator was called!"
            #do stuff after
        return decorator

    @abstractmethod
    def inherited():
        raise NotImplemented


class B(A):
    def __init__(self):
        super(B,self).__init__()

    @A.the_decorator #<--- this is what I want, 
    @overrides
    #@A.the_decorator <--- or this
    def inherited(self):
        print "B.inherited was invoked"

我可以在 Python 2.7 下测试它，除了 @overrides 装饰器(我在测试中评论了它)