Guido van Rossum wrote:
>>Even better would to have "X.Y.__outerclass__ is X",
>>i.e. __outerclass__ as a (weak) reference to the class
>>in which X.Y was defined. What I came up with is this:
>>
>>class nestedtype(type):
>> def __new__(cls, name, bases, dict):
>> dict["__outerclass__"] = None
>> res = type.__new__(cls, name, bases, dict)
>> for (key, value) in dict.items():
>> if isinstance(value, type):
>> value.__outerclass__ = res
>> return res
>>
>> def __fullname__(cls):
>> name = cls.__name__
>> while 1:
>> cls = cls.__outerclass__
>> if cls is None:
>> return name
>> name = cls.__name__ + "." + name
>>
>> def __repr__(cls):
>> return "<class %s/%s at 0x%x>" % \
>> (cls.__module__, cls.__fullname__(), id(cls))
>
>
> I kind of like __fullname__ as a useful attribute (though I'd make it
> an attribute or property [== computed attribute] rather than a
> method). But if setting the __name__ of an inner class to
> "Outer.Inner" doesn't break too much existing code, I'd prefer that --
> no new attributes needed.
I guess this shouldn't be a problem.
> I don't see any use for __outerclass__ *except* for computing the full
> name. And there are better ways to do that if you have help from the
> parser.
Exactly, I only used __outerclass__, because I didn't have the help
of the parser and couldn't think of any other way to implement it.
What I wonder is how this will work with classes that are defined
outside but assigned inside an outer class, i.e.:
class NotInner:
pass
class Outer:
Inner = NotInner
Will this set NotInner.__name__ to "Outer.NotInner" or not?
Bye,
Walter Dörwald
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