"GvR" == Guido van Rossum email@example.com writes:
GvR> Let x be an object, C its class, and M C's class. So, | x.__class__ is C | C.__class__ is M GvR> Then x's methods are described in C.__dict__, and C's methods GvR> are described in M.__dict__. GvR> The problem is that if you write C.spam, there could be two GvR> spams: one in C.__dict__, one in M.__dict__. Which one to GvR> use?
If you use naming to generally distinguish, and have a lookup chain that first found it in C.__dict__ and then looked in M.__dict__, you could control what happens when the name is in both dicts by using a more explicit lookup, e.g. C.__dict__['meth'] vs. C.__class__.__dict__['meth']
Couldn't be C.__class__.meth be used?
But maybe that's too ugly. GvR> How does Smalltalk resolve this?
I'm walking on thin ice here (maybe I should better try it out), but IIRC Smalltalk requires to explicit:
self class method; or self method;
Another question: presumably when I write
class Foo: pass
Foo is implicitly given the built-in metaclass M, but say I wanted to define a class Foo with a different metaclass, how would I spell this? I think at one point I suggested a semi-ugly syntactic hack, where `class' was actually a namespace and you could add new metaclasses to it. So you could write something like
class.WeirdClass Foo: pass
and now Foo's metaclass would be WeirdClass.
Thin ice again I'm on here (even more), but I have the impression that creating a class Point in Smalltalk _automatically_ creates two classes: Point and PointClass. The latter is normally hidden (but contains the class methods of Point as instance methods).