Can I find the class of a method in a decorator.

[Chris Angelico]

On Sun, Mar 6, 2016 at 2:05 AM, Antoon Pardon
<antoon.pardon at rece.vub.ac.be> wrote:
> Using python 3.4/3.5
>
> Suppose I have the following class:
>
> class Tryout:
>
>     @extern
>     def method(self, ...)
>
> Now how can I have access to the Tryout class in
> the extern function when it is called with method
> as argument
>
> def extern(f):
>     the_class = ????
>
> f.__class doesn't work, if I write the following
>
> def extern(f)
>     print(f.__class__)
>
> the result is: <class 'function'>, so that doesn't work.
> Looking around I didn't find an other obvious candidate
> to try. Anybody an idea?

At the time when the function decorator is run, there isn't any class.
You could just as effectively create your function outside the class
and then inject it (Tryout.method = method).

What is it you're trying to do? Would it be a problem to have a class
decorator instead/as well?

ChrisA