Actually i was not mutable. Try this: i = 1 id(i) i += 1 id(i)
Change both of your sample functions to also print the id of the default variable and that might shed a little more light on the matter. "i = i + 1" is only changing the local reference called i to refer to an instance of a different integer object. If you change the original function to something similar, it will behave similarly: def f( a, L=[]): L = L + [a] return L Because you are assigning the local reference variable L to a new list, instead of modifying that original default list that was created. Is that more clear, or is it now more unclear? ;) -- http://mail.python.org/mailman/listinfo/python-list
