How to find class of bound method during class construction in Python 3.1?

Question

i want to write a decorator that enables methods of classes to become visible to other parties; the problem i am describing is, however, independent of that detail. the code will look roughly like this:

def CLASS_WHERE_METHOD_IS_DEFINED( method ):
  ???

def foobar( method ):
  print( CLASS_WHERE_METHOD_IS_DEFINED( method ) )

class X:

  @foobar
  def f( self, x ):
    return x ** 2

my problem here is that the very moment that the decorator, foobar(), gets to see the method, it is not yet callable; instead, it gets to see an unbound version of it. maybe this can be resolved by using another decorator on the class that will take care of whatever has to be done to the bound method. the next thing i will try to do is to simply earmark the decorated method with an attribute when it goes through the decorator, and then use a class decorator or a metaclass to do the postprocessing. if i get that to work, then i do not have to solve this riddle, which still puzzles me:

can anyone, in the above code, fill out meaningful lines under CLASS_WHERE_METHOD_IS_DEFINED so that the decorator can actually print out the class where f is defined, the moment it gets defined? or is that possibility precluded in python 3?

Answer 1

When the decorator is called, it's called with a function as its argument, not a method -- therefore it will avail nothing to the decorator to examine and introspect its method as much as it wants to, because it's only a function and carries no information whatsoever about the enclosing class. I hope this solves your "riddle", although in the negative sense!

Other approaches might be tried, such as deep introspection on nested stack frames, but they're hacky, fragile, and sure not to carry over to other implementations of Python 3 such as pynie; I would therefore heartily recommend avoiding them, in favor of the class-decorator solution that you're already considering and is much cleaner and more solid.