Question

假设我有一个班级

class SimpleGenerator(object):
    @classmethod
    def get_description(cls):
        return cls.name

class AdvancedGenerator(SimpleGenerator):
    @classmethod
    def get_description(cls):
        desc = SimpleGenerator.get_description() # this fails
        return desc + ' Advanced(tm) ' + cls.adv_feature

现在我已经扩展了上面的每一个类，每个类都有一个具体的类：

class StringGenerator(SimpleGenerator)
    name = 'Generates strings'
    def do_something():
        pass

class SpaceShuttleGenerator(AdvancedGenerator)
    name = 'Generates space shuttles'
    adv_feature = ' - builds complicated components'
    def do_something():
        pass

现在让我说我打电话

SpaceShuttleGenerator.get_description()

问题在于AdvancedGenerator我希望调用SimpleGenerator中的方法传递类的实例，特别是SpaceShuttleGenerator。可以这样做吗？

注意：该示例已简化，因为我的具体示例更为复杂。让我们说我的目标不是连接字符串。

Answer 1

使用super()：

@classmethod
def get_description(cls):
    desc = super(AdvancedGenerator, cls).get_description()
    return desc + ' Advanced(tm) ' + cls.adv_feature

使用SimpleGenerator.get_description()和super(AdvancedGenerator, cls).get_description()之间的区别在于cls将被设置为。直接调用课程时，cls设置为SimpleGenerator，使用super()，cls会引用AdvancedGenerator。

比较您的代码（调整为使用__name__来说明差异）：

>>> class SimpleGenerator(object):
...     @classmethod
...     def get_description(cls):
...         return cls.__name__
... 
>>> class AdvancedGenerator(SimpleGenerator):
...     @classmethod
...     def get_description(cls):
...         desc = SimpleGenerator.get_description() 
...         return desc + ' Advanced(tm)'
... 
>>> AdvancedGenerator.get_description()
'SimpleGenerator Advanced(tm)'

并使用super()：

>>> class AdvancedGenerator(SimpleGenerator):
...     @classmethod
...     def get_description(cls):
...         desc = super(AdvancedGenerator, cls).get_description()
...         return desc + ' Advanced(tm)'
... 
>>> AdvancedGenerator.get_description()
'AdvancedGenerator Advanced(tm)'

如何从Python中重写的@classmethod调用父类的@classmethod？

1 个答案: