Overriding a static method in python
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).
Solution 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() # -> 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
Solution 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()
.