Question

I’ve read in Expert Python Programming about this edge case. Check this code:

def f(arg={}):
    arg['3'] = 4
    return arg

>>> print f()
{'3': 4}
>>> res = f()
>>> res['4'] = 'Still here'
>>> print f()
{'3': 4, '4': 'Still here'}

It’s not clear to me why when f gets called the last time (after its return value has been saved), instead of assigning arg the empty dict (since it was called with no arguments), it keeps the old reference.

The book says so: “if an object is created within the arguments, the argument reference will still be alive if the function returns the object”.

I understand that “this is the way it works”, but why so?

Answer 1

Because the default arguments are evaluated once only, when the function is evaluated and created (they are part of the function defenition and can be fetched through inspect.getargspec, for example).

Since they are part of the function, every call to the function will have the same instance of the default value. This is not a problem if it is an immutable value, but as soon as it is mutable it can become a gotcha.

The same ‘feature’ exist in class defenitions, given a class defenition:

class A(object):
    foo = {}

calling

x = A() 
y = A()
x.foo['bar'] = "baz"

…would give that y.foo[‘bar’] evaluates to “baz”, since x and y has the same foo.
This is why member initialization should be done in init instead of the class body.