Python Method overriding, does signature matter?

Question

Lets say I have

class Super():
  def method1():
    pass

class Sub(Super):
  def method1(param1, param2, param3):
      stuff

Is this corr

Answer 1

It will work:

>>> class Foo(object):
...   def Bar(self):
...     print 'Foo'
...   def Baz(self):
...     self.Bar()
... 
>>> class Foo2(Foo):
...   def Bar(self):
...     print 'Foo2'
... 
>>> foo = Foo()
>>> foo.Baz()
Foo
>>> 
>>> foo2 = Foo2()
>>> foo2.Baz()
Foo2

However, this isn't generally recommended. Take a look at S.Lott's answer: Methods with the same name and different arguments are a code smell.

Answer 2

In Python, methods are just key-value pairs in the dictionary attached to the class. When you are deriving a class from a base class, you are essentially saying that method name will be looked into first derived class dictionary and then in the base class dictionary. In order to "override" a method, you simply re-declare the method in the derived class.

So, what if you change the signature of the overridden method in the derived class? Everything works correctly if the call is on the derived instance but if you make the call on the base instance, you will get an error because the base class uses a different signature for that same method name.

There are however frequent scenarios where you want derived class method have additional parameters and you want method call work without error on base as well. This is called "Liskov substitution principle" (or LSP) which guarantees that if person switches from base to derived instance or vice versa, they don't have to revamp their code. To do this in Python, you need to design your base class with the following technique:

class Base:
    # simply allow additional args in base class
    def hello(self, name, *args, **kwargs):
        print("Hello", name)

class Derived(Base):
      # derived class also has unused optional args so people can
      # derive new class from this class as well while maintaining LSP
      def hello(self, name, age=None, *args, **kwargs):
          super(Derived, self).hello(name, age, *args, **kwargs) 
          print('Your age is ', age)

b = Base()
d = Derived()

b.hello('Alice')        # works on base, without additional params
b.hello('Bob', age=24)  # works on base, with additional params
d.hello('Rick')         # works on derived, without additional params
d.hello('John', age=30) # works on derived, with additional params

Above will print:

    Hello Alice
    Hello Bob
    Hello Rick
    Your age is  None
    Hello John
    Your age is  30

. Play with this code

Answer 3

Python will allow this, but if method1() is intended to be executed from external code then you may want to reconsider this, as it violates LSP and so won't always work properly.

Answer 4

In python, all class methods are "virtual" (in terms of C++). So, in the case of your code, if you'd like to call method1() in super class, it has to be:

class Super():
    def method1(self):
        pass

class Sub(Super):
    def method1(self, param1, param2, param3):
       super(Sub, self).method1() # a proxy object, see http://docs.python.org/library/functions.html#super
       pass

And the method signature does matter. You can't call a method like this:

sub = Sub()
sub.method1() 

Answer 5

Yes. Calls to "method1" will always go to the subclass. Method signature in Python only consist of the name and not the argument list.

Answer 6

You could do something like this if it's ok to use default arguments:

>>> class Super():
...   def method1(self):
...     print("Super")
...
>>> class Sub(Super):
...   def method1(self, param1="X"):
...     super(Sub, self).method1()
...     print("Sub" + param1)
...
>>> sup = Super()
>>> sub = Sub()
>>> sup.method1()
Super
>>> sub.method1()
Super
SubX