Using a mutable default argument, how to access the object from outside the function?

Question

If I have something like this:

def f(x, cache=[]):
    cache.append(x)

How can I access cache from outside f?

Answer 1

You can use __defaults__ magic:

>>> def f(x, cache=[]):
...     cache.append(x)
>>> f.__defaults__
([],)
>>> f(2)
>>> f.__defaults__
([2],)
>>> f('a')
>>> f.__defaults__
([2, 'a'],)
>>> c, = f.__defaults__
>>> c
[2, 'a']

Just for the sake of completeness, inspect.getfullargspec can also be used, which is more explicit:

>>> import inspect
>>> inspect.getfullargspec(f)
FullArgSpec(args=['x', 'cache'], varargs=None, varkw=None, defaults=([2, 'a'],), kwonlyargs=[], kwonlydefaults=None, annotations={})
>>> inspect.getfullargspec(f).defaults
([2, 'a'],)