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在Python中调用基类的类方法

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问题：在Python中调用基类的类方法
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回答 1
回答 2

问题：在Python中调用基类的类方法

考虑以下代码：

class Base(object):

    @classmethod
    def do(cls, a):
        print cls, a

class Derived(Base):

    @classmethod
    def do(cls, a):
        print 'In derived!'
        # Base.do(cls, a) -- can't pass `cls`
        Base.do(a)

if __name__ == '__main__':
    d = Derived()
    d.do('hello')

> $ python play.py  
> In derived! 
> <class '__main__.Base'> msg

从哪里来Derived.do，我怎么打电话Base.do？

super如果这是一个普通的对象方法，通常我会直接使用甚至直接使用基类名称，但是显然我找不到在基类中调用类方法的方法。

在上面的示例中，Base.do(a)打印Baseclass而不是Derivedclass。

点击查看英文原文

Consider the following code:

class Base(object):

    @classmethod
    def do(cls, a):
        print cls, a

class Derived(Base):

    @classmethod
    def do(cls, a):
        print 'In derived!'
        # Base.do(cls, a) -- can't pass `cls`
        Base.do(a)

if __name__ == '__main__':
    d = Derived()
    d.do('hello')

> $ python play.py  
> In derived! 
> <class '__main__.Base'> msg

From Derived.do, how do I call Base.do?

I would normally use super or even the base class name directly if this is a normal object method, but apparently I can’t find a way to call the classmethod in the base class.

In the above example, Base.do(a) prints Base class instead of Derived class.

回答 0

如果您使用的是新样式的类（例如，是从objectPython 2 派生的，或者总是在Python 3中派生的），则可以这样操作super()：

super(Derived, cls).do(a)

这是您如何在基类版本的方法（即print cls, a）中从派生类调用代码并将cls其设置为派生类的方式。

点击查看英文原文

If you’re using a new-style class (i.e. derives from object in Python 2, or always in Python 3), you can do it with super() like this:

super(Derived, cls).do(a)

This is how you would invoke the code in the base class’s version of the method (i.e. print cls, a), from the derived class, with cls being set to the derived class.

回答 1

这已经有一段时间了，但是我想我可能已经找到了答案。当您装饰一个方法成为类方法时，原始的未绑定方法存储在名为“ im_func”的属性中：

class Base(object):
    @classmethod
    def do(cls, a):
        print cls, a

class Derived(Base):

    @classmethod
    def do(cls, a):
        print 'In derived!'
        # Base.do(cls, a) -- can't pass `cls`
        Base.do.im_func(cls, a)

if __name__ == '__main__':
    d = Derived()
    d.do('hello')
点击查看英文原文

this has been a while, but I think I may have found an answer. When you decorate a method to become a classmethod the original unbound method is stored in a property named ‘im_func’:

class Base(object):
    @classmethod
    def do(cls, a):
        print cls, a

class Derived(Base):

    @classmethod
    def do(cls, a):
        print 'In derived!'
        # Base.do(cls, a) -- can't pass `cls`
        Base.do.im_func(cls, a)

if __name__ == '__main__':
    d = Derived()
    d.do('hello')

回答 2

这对我有用：

Base.do('hi')
点击查看英文原文

This works for me:

Base.do('hi')