Question

I am trying to dynamically assign a function to __del__ of an instance of a class so that it gets called when using dir() on that object. I need __dir__ to be unique for each instance of the class. As a stripped down example, I have tried:

import types
class Foo(object):
    def __init__(self, arg):
        def __dir__(self):
            print "in __dir__"
            return [arg]
        self.__dir__ = types.MethodType(__dir__, self, self.__class__)

foo = Foo('bar')
print dir(foo)
print
print foo.__dir__()

This prints:

['__class__', '__delattr__', '__dict__', '__dir__', '__doc__', '__format__', '__getattribute__', '__hash__', '__init__', '__module__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__sizeof__', '__str__', '__subclasshook__', '__weakref__']

in __dir__
['bar']

If I instead do this:

class Foo(object):
    def __dir__(self):
        print "in __dir__"
        return ['bar']

foo = Foo()
print dir(foo)

that outputs:

in __dir__
['bar']

as expected, but can not be customized for each instance of the class. Any ideas?

Answer 1

Ok, based on your comment I worked things a bit. I think this might be closer to the behavior you’re looking for. Calling dir with no args gives all the names in the local scope and calling it on __class__ gives all class names. This ignores the defined __dir__ which can be called later on. I’m curious what you’re using this for, maybe there is a simpler way to get the intended behavior?

class Foo(object):
    def __init__(self, arg=None):
        self.arg = arg
        print dir(self.__class__) + dir()

    def __dir__(self):
        return [self.arg()]


def dirfoo():
    return ["new thing"]

foo = Foo(dirfoo)

print dir(foo)