Overriding a static method in python

Question

Hello,

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()

produces output:

Called method_one
Called method_two
Called method_two
Called method_two

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

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

Answer 1

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()
a_test.method_two()
a_test.method_three()
b_test = T2()
b_test.method_three()

Test.method_two()

T2.method_three()

Answer 2

The behavior you see is the expected behavior. Static methods are... static. When you call method_three() defined in Test it will certainly call method_two() defined by Test.

As for how to "get around" this proper behavior...

The very best way is to make methods virtual when you want virtual behavior. If you're stuck with some library code with a static method that you wish were virtual then you might look deeper to see if there's a reason or if it's just an oversight.

Otherwise, you can define a new method_three() in T2 that calls T2.method_two().