How do I give a decorator acces to the class of a decorated function

Thu Sep 5 06:13:33 EDT 2019

On 4/09/19 17:46, Peter Otten wrote:
> Antoon Pardon wrote:
>>> What I am trying to do is the following.
>>>> class MyClass (...) :
>> @register
>> def MyFunction(...)
>> ...
>>>> What I would want is for the register decorator to somehow create/mutate
>> class variable(s) of MyClass.
>>>> Is that possible or do I have to rethink my approach?
> If you are willing to delegate the actual work to the metaclass call: 
>> def register(f):
> f.registered = True
> return f
>> def registered(name, bases, namespace):
> namespace["my_cool_functions"] = [
> n for n, v in namespace.items()
> if getattr(v, "registered", False)
> ]
> return type(name, bases, namespace)
>> class MyClass(metaclass=registered) :
> @register
> def foo(self):
> pass
> @register
> def bar(self):
> pass
> def other(self):
> pass
>> print(MyClass.my_cool_functions)

I have been playing with this idea and it looks promising. I was wondering
about two points.
1) I guess I can add extra methods to my class through the metaclass by
 having something like the following in the registered function:
def registered(name, bases, namespace):
 namespace["my_cool_functions"] = [
 n for n, v in namespace.items()
 if getattr(v, "registered", False)
 ]
 namespace["__getitem__"] = lambda self, index: self.my_cool_functions[index]
2) Is it possible to make MyClass automatically a subclass of an other class
 through the metaclass?
-- 
Antoon.