On Sun, 21 Feb 2010 04:50:49 am Alan Harris-Reid wrote:
> Hi,
>
> I am having trouble understanding how superclass calls work. Here's
> some code...
What version of Python are you using?
In Python 2.x, you MUST inherit from object to use super, and you MUST
explicitly pass the class and self:
class ParentClass(object):
def __init__(self, a, b, c):
do something here
class ChildClass(ParentClass):
def __init__(self, a, b, c):
super(ChildClass, self).__init__(a, b, c)
In Python 3.x, all classes inherit from object and you no longer need to
explicitly say so, and super becomes a bit smarter about where it is
called from:
# Python 3 only
class ParentClass:
def __init__(self, a, b, c):
do something here
class ChildClass(ParentClass):
def __init__(self, a, b, c):
super().__init__(a, b, c)
I assume you are using Python 3.0 or 3.1. (If you're 3.0, you should
upgrade to 3.1: 3.0 is s-l-o-w and no longer supported.)
Your mistake was to pass self as an explicit argument to __init__. This
is not needed, because Python methods automatically get passed self:
> def __init__(self):
> super().__init__(self)
That has the effect of passing self *twice*, when __init__ expects to
get self *once*, hence the error message you see:
> When the super().__init__ line runs I get the error "__init__() takes
> exactly 1 positional argument (2 given)"
--
Steven D'Aprano
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