在python中重写静态方法 [英] Overriding a static method in python

问题描述

请参考第一个答案关于python的绑定和未绑定方法，我有一个问题:

Referring to the first answer about python's bound and unbound methods here, I have a question:

class Test:
    def method_one(self):
        print "Called method_one"
    @staticmethod
    def method_two():
        print "Called method_two"
    @staticmethod
    def method_three():
        Test.method_two()
class T2(Test):
    @staticmethod
    def method_two():
        print "T2"
a_test = Test()
a_test.method_one()
a_test.method_two()
a_test.method_three()
b_test = T2()
b_test.method_three()

产生输出:

Called method_one
Called method_two
Called method_two
Called method_two

有没有一种方法可以覆盖python中的静态方法?

Is there a way to override a static method in python?

我希望b_test.method_three()打印"T2"，但不会(打印"Called method_two").

I expected b_test.method_three() to print "T2", but it doesn't (prints "Called method_two" instead).

推荐答案

在此处使用的形式中，您明确指定了要调用的类的静态method_two.如果method_three是类方法，并且您调用了cls.method_two，则将获得所需的结果:

In the form that you are using there, you are explicitly specifying what class's static method_two to call. If method_three was a classmethod, and you called cls.method_two, you would get the results that you wanted:

class Test:
    def method_one(self):
        print "Called method_one"
    @staticmethod
    def method_two():
        print "Called method_two"
    @classmethod
    def method_three(cls):
        cls.method_two()

class T2(Test):
    @staticmethod
    def method_two():
        print "T2"

a_test = Test()
a_test.method_one()  # -> Called method_one
a_test.method_two()  # -> Called method_two
a_test.method_three()  # -> Called method_two

b_test = T2()
b_test.method_three()  # -> T2
Test.method_two()  # -> Called method_two
T2.method_three()  # -> T2

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