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<blockquote cite="mid:mailman.48.1266722371.4576.tutor@python.org"
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<pre wrap="">Hi,
I am having trouble understanding how superclass calls work. Here's
some code...
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<pre wrap=""><!---->
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:
</pre>
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<pre wrap=""> def __init__(self):
super().__init__(self)
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<pre wrap=""><!---->
That has the effect of passing self *twice*, when __init__ expects to
get self *once*, hence the error message you see:
</pre>
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<pre wrap="">When the super().__init__ line runs I get the error "__init__() takes
exactly 1 positional argument (2 given)"
</pre>
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</blockquote>
Hi Steven, thanks for the reply.<br>
<br>
Fortunately I am using Python 3.1, so I can use the super().__init__(a,
b, c) syntax.<br>
<br>
Regards,<br>
Alan<br>
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