Does dict.update affect a function's argspec?

Question

import inspect
class Test:
  def test(self, p, d={}):
    d.update(p)
    return d
print inspect.getargspec(getattr(Test, 'test'))[3]
print Test().test({'1':True})
print inspect.getargspec(getattr(Test, 'test'))[3]

I would expect the argspec for Test.test not to change but because of dict.update it does. Why?

Answer 1

Because dicts are mutable objects. When you call d.update(p), you are actually mutating the default instance of the dict. This is a common catch; in particular, you should never use a mutable object as a default value in the list of arguments.

A better way to do this is as follows:

class Test:
    def test(self, p, d = None):
        if d is None:
            d = {}
        d.update(p)
        return d

Answer 2

A default argument in Python is whatever object was set when the function was defined, even if you set a mutable object. This question should explain what that means and why Python is the SO question least astonishment in python: the mutable default argument.

Basically, the same default object is used every time the function is called, rather than a new copy being made each time. For example:

>>> def f(xs=[]):
...   xs.append(5)
...   print xs
... 
>>> f()
[5]
>>> f()
[5, 5]

The easiest way around this is to make your actual default argument None, and then simply check for None and provide a default in the function, for example:

>>> def f(xs=None):
...   if xs is None:
...     xs = []
...   xs.append(5)
...   print xs
... 
>>> f()
[5]
>>> f()
[5]