How do I initialize the base (super) class?


Question

In Python, consider I have the following code:

>>> class SuperClass(object):
    def __init__(self, x):
        self.x = x

>>> class SubClass(SuperClass):
    def __init__(self, y):
        self.y = y
        # how do I initialize the SuperClass __init__ here?

How do I initialize the SuperClass __init__ in the subclass? I am following the Python tutorial and it doesn't cover that. When I searched on Google, I found more than one way of doing. What is the standard way of handling this?

1
111
12/26/2016 10:48:47 PM

Accepted Answer

Python (until version 3) supports "old-style" and new-style classes. New-style classes are derived from object and are what you are using, and invoke their base class through super(), e.g.

class X(object):
  def __init__(self, x):
    pass

  def doit(self, bar):
    pass

class Y(X):
  def __init__(self):
    super(Y, self).__init__(123)

  def doit(self, foo):
    return super(Y, self).doit(foo)

Because python knows about old- and new-style classes, there are different ways to invoke a base method, which is why you've found multiple ways of doing so.

For completeness sake, old-style classes call base methods explicitly using the base class, i.e.

def doit(self, foo):
  return X.doit(self, foo)

But since you shouldn't be using old-style anymore, I wouldn't care about this too much.

Python 3 only knows about new-style classes (no matter if you derive from object or not).

133
3/19/2018 5:27:27 PM

Both

SuperClass.__init__(self, x)

or

super(SubClass,self).__init__( x )

will work (I prefer the 2nd one, as it adheres more to the DRY principle).

See here: http://docs.python.org/reference/datamodel.html#basic-customization


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