Wrong syntax when nesting 3 classes in Python?

Question

Would there be a simple way to fix this error while keeping all 3 levels?

Deriving ClassA from object does not help.

Thanks in advance!

>>> class classA:
... class classB(object):
... def __init__(self):
... self.b = 3
... class classC(classA.classB):
... def __init__(self):
... super(classC, self).__init__()
... self.c = 4
...
Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
 File "<stdin>", line 2, in classA
 File "<stdin>", line 5, in classB
NameError: name 'classA' is not defined

• Daniel Roseman: Why would you want to do this?
• Yulia V: class names will be shorter and code will be more readable
• Yulia V: also, wanted to understand why thid=s does not work; maybe I miss some important concept
• glglgl: classA.classB.classC is no way shorter than classC...
• Markus Unterwaditzer: @glglgl Maybe he saves one or another prefix with this.

Answer 1

No. At the time you define classC, classA does not exist yet.

It is only created after its body is fully executed. (The dict created from the body's execution is one parameter for the class creation call class = type('classname', (superclass_1, superclass_2, superclass_3), said_dict}).)

The easiest way would be defining them at the same level.

If absolutely needed, you can modify them later:

class classA:
 pass
class classB(object):
 def __init__(self):
 self.b = 3
class classC(classB):
 def __init__(self):
 super(classC, self).__init__()
 self.c = 4
classA.classB = classB
classB.classC = classC

• Duncan: It might be useful to add to your answer an explanation of why classA cannot exist until all of the statements in the class body have been executed.

Answer 2

Maybe something like this would work:

class classA:
 pass
class classB(object):
 def __init__(self):
 self.b = 3
classA.classB = classB
class classC(classA.classB):
 def __init__(self):
 super(classC, self).__init__()
 self.c = 4
classA.classB.classC = classC

• Duncan: Apart from makeing the last line more verbose, this doesn't add anything to @glglgl's answer.
• Facundo Casco: @Duncan, you are right but when I posted it glglgl's answer didn't have the code