How to get class when I can't use self - Python

Question

I have one weird problem. I have following code:

class A:
    def f():
        return __class__()

class B(A):
    pass
a = A.f()
b = B.f()
print(a, b)

And output is something like this:

<__main__.A object at 0x01AF2630> <__main__.A object at 0x01B09B70>

So how can I get B instead of second A?

`how can I get B instead of second A` - What do you mean by that? An instance of `B`? — thefourtheye, Apr 18 '15 at 11:48
In printed text. What I meant is that `b` becomes instance of `B`. — knowledge, Apr 18 '15 at 12:20

Martijn Pieters · Accepted Answer · 2015-04-18T11:58:25.543

7

The magic __class__ closure is set for the method context and only really meant for use by super().

For methods you'd want to use self.__class__ instead:

return self.__class__()

or better still, use type(self):

return type(self)()

If you want to be able to call the method on a class, then use the classmethod decorator to be handed a reference to the class object, rather than remain unbound:

@classmethod
def f(cls):
    return cls()

classmethods are always bound to the class they are called on, so for A.f() that'd be A, for B.f() you get handed in B.

edited Apr 18 '15 at 11:58

answered Apr 18 '15 at 11:42

Martijn Pieters

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@thefourtheye: right, then using a `@classmethod` is the right way to go about it. – Martijn Pieters Apr 18 '15 at 11:47
`@classmethod` is excellent, I did't use `self.__class__()` because I couldn't use `self`. I always thought `@classmethod` use object not class. – knowledge Apr 18 '15 at 11:57

How to get class when I can't use self - Python

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