This actually has nothing to do with default values, other than that it often comes up as an unexpected behaviour when you write functions with mutable default values.
>>> def foo(a):
a.append(5)
print a
>>> a = [5]
>>> foo(a)
[5, 5]
>>> foo(a)
[5, 5, 5]
>>> foo(a)
[5, 5, 5, 5]
>>> foo(a)
[5, 5, 5, 5, 5]
No default values in sight in this code, but you get exactly the same problem.
The problem is that foo
is modifying a mutable variable passed in from the caller, when the caller doesn't expect this. Code like this would be fine if the function was called something like append_5
; then the caller would be calling the function in order to modify the value they pass in, and the behaviour would be expected. But such a function would be very unlikely to take a default argument, and probably wouldn't return the list (since the caller already has a reference to that list; the one it just passed in).
Your original foo
, with a default argument, shouldn't be modifying a
whether it was explicitly passed in or got the default value. Your code should leave mutable arguments alone unless it is clear from the context/name/documentation that the arguments are supposed to be modified. Using mutable values passed in as arguments as local temporaries is an extremely bad idea, whether we're in Python or not and whether there are default arguments involved or not.
If you need to destructively manipulate a local temporary in the course of computing something, and you need to start your manipulation from an argument value, you need to make a copy.