Question

Python Tutorial –

Important warning: The default value is evaluated only once
…
If you don’t want the default to be
shared between subsequent calls, you
can write the function like this
instead:

def f(a, L=None):
    if L is None:
        L = []
    L.append(a)
    return L

I still was expecting:

print f(1)    # [1]
print f(2)    # [1,2]
print f(3)    # [1,2,3]

I reason:

The default value (L=None) was executed for f(1) which helped L point to a new empty list in the fn body. However on successive calls, L=None was not executed; so L still points to the list which already has 1 in it now, and subsequent calls are simply appending more elements to it thereby sharing L.

Where am I thinking incorrectly?

---

UPDATE

def f(a, L=[]):
    L.append(a)
    return L

Does L here point to an empty list created in heap or stack?

Answer 1

L is the name of an argument, but it is also a local variable. Rebinding it rebinds the local variable, but does not change the default argument.

UPDATE EDIT:

Python doesn’t have “heap” and “stack” in the same manner as C; all it has are objects, and references to those objects. Your function call returns a reference to the same list that was created as the default value for the L argument of the f function, and any operation that mutates it will mutate the value of the default argument.