class A:
class B(A):
pass
This does not work. Why? How do I make something that behaves the same way?
By same I mean:
- The same
__class__.__name__ - The same
__class__.__qualname__ A.B.__base__ is A
3 Answers
No, not without workarounds. The first class doesn't even exist at the time its body (including the class B(A) part) is executed. You can define the two separtely though:
class A:
pass
class B(A):
pass
A.B = B
B.__name__ = ...
B.__qualname__ = ...
But honestly this doesn't look like the best design or API.
Comments
Evaluation happens "inside out". Class A is not created until after everything in its class body is done. So at the time B is defined, A does not yet exist.
Your only option would be to define class B with some other base class, then write a class decorator or metaclass on A that looks inside A for attributes that are classes with the specified "dummy" base, and recreates them with A as their base.
The real question is why you feel you need to do this.
Comments
Just because metaclasses are amusing:
class GoofyMeta(type):
def __new__(cls, clsname, bases, dct):
dct['__qualname__'] = bases[0].__qualname__
return super(GoofyMeta, cls).__new__(cls, bases[0].__name__, bases, dct)
class A:
pass
#using python 3 syntax here
class B(A,metaclass=GoofyMeta):
pass
B.__name__
Out[24]: 'A'
B.__qualname__
Out[25]: 'A'
And B.__base__ is A by definition.
As for why you'd want this, I'm pretty mystified. It seems like you can just use inheritance/composition normally like the rest of us :-)
(I'll also note that I didn't address B somehow being an inner class of A. I extra plus one don't know what you're trying to do with that.)
name A not defined, it's because the definition ofAisn't done by the timeBis beginning to get defined.line 2: NameError: name 'A' is not defined.