CODEWITHSUNDEEP
python
Prem Anand
Prem Anand

Reputation: 2537

Error on using super: __init__() takes exactly 1 argument

Can anyone explain why I am getting __init__() takes exactly 1 argument error with the below code

class ABC(object):
    def __init__(self, **kwargs):
        self.x=20
        super(ABC, self).__init__(self, **kwargs)

class DEF(object):
     def __init__(self, **kwargs):
        self.y=30
        super(DEF, self).__init__(self, **kwargs)

class XYZ(ABC,DEF):
     def __init__(self, **kwargs):
        super(XYZ, self).__init__(self, **kwargs)

>>> x=XYZ()
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/home/user/test.py", line 13, in __init__
    super(XYZ, self).__init__(self, **kwargs)
TypeError: __init__() takes exactly 1 argument (2 given)

>>> x=XYZ(a=1)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/home/hprem/junk", line 13, in __init__
    super(XYZ, self).__init__(self, **kwargs)
TypeError: __init__() takes exactly 1 argument (3 given)

Upvotes: 0

Views: 4543

Answers (1)

BrenBarn
BrenBarn

Reputation: 251365

You don't need to pass self explicitly to the superclass __init__. By doing super(XYZ, self) you created a "bound" super object that "knows" what self is. Just do super(XYZ, self).__init__(**kwargs) (and likewise for your other calls).

Note, though, that this class structure will fail if you pass any kwargs. object.__init__ doesn't accept any parameters, so you can't pass it any. For this reason, a class that inherits directly from object can't safely use super in general (unless it explicitly calls it with no arguments, in which case there's no point in calling it at all). It's better to create a "topmost" superclass from which ABC and DEF inherit.

Upvotes: 7

Related Questions