Using super within nested class

Question

I am puzzled with the following error in python 2.7.12 Suppose we have a class definition within a class, something similar to this:

class C(object):
    def __init__(self):
        print "class C"


class D(object):
    def __init__(self):
        print "class D"


class A(D):

    class B(C):
        def __init__(self):
            # Strangely here B is "not defined", why?
            super(B, self).__init__()
            print "class B"

    def __init__(self):
        super(D, self).__init__()
        print "class A"

    def do_something(self):
        b_class = self.B()
        print "b_class within A : {}".format(b_class)


a_class = A()
a_class.do_something()

but if we we extract the definition of class B outside the scope of class A, everything works well.

Do we need to use "super" differently when called within a nested class? I fail to understand why its usage would be different within or outside the nested class. Any pointers?

Any pointers? There are no pointers in Python ;)

Leon
– Leon

2016-11-07 14:00:45 +00:00
Commented Nov 7, 2016 at 14:00 — Leon
– Leon, Commented Nov 7, 2016 at 14:00

Daniel Roseman · Accepted Answer · 2016-11-07 13:31:06Z

7

The problem is not the subclass or superclass, but the nesting. B itself is not defined, only A.B is.

Note that in Python there is almost never a good reason to nest classes, though.

answered Nov 7, 2016 at 13:31

Daniel Roseman

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1 Comment

TocToc Over a year ago

Indeed that is the case, although I would have thought that super implicitly took the context into account...

jonrsharpe · Accepted Answer · 2016-11-07 13:58:46Z

4

You need to address B by its full name, A.B:

class C(object):
    def __init__(self):
        print "class C"


class D(object):
    def __init__(self):
        print "class D"


class A(D):

    class B(C):
        def __init__(self):
            super(A.B, self).__init__()
            print "class B"

    def __init__(self):
        super(D, self).__init__()
        print "class A"

    def do_something(self):
        b_class = self.B()
        print "b_class within A : {}".format(b_class)


>>> a_class = A()
>>> a_class.do_something()

class A
class C
class B
b_class within A : <__main__.B object at 0x7f0cac98cbd0>

edited Nov 7, 2016 at 13:58

jonrsharpe

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answered Nov 7, 2016 at 13:31

Sebastian Wozny

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1 Comment

bruno desthuilliers Over a year ago

@Leon that's because A.__init__ calls super(D, self).__init__(), which means "calls the __init__ of D superclass" (in this case object.__init__). If you want to invoke D.__init__ from A.__init__, you have to use super(A, self).__init__() instead ;)

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