The default arguments are evaluated once when the function is defined.
So you get the same list object each time the function is called.
You'll also get the same 0 object each time the second function is called, but since int is immutable, when you add 1 as fresh object needs to be bound to a
>>> def foo(L = []):
... print id(L)
... L.append(1)
... print id(L)
... print L
...
>>> foo()
3077669452
3077669452
[1]
>>> foo()
3077669452
3077669452
[1, 1]
>>> foo()
3077669452
3077669452
[1, 1, 1]
vs
>>> def foo(a=0):
... print id(a)
... a+=1
... print id(a)
... print a
...
>>> foo()
165989788
165989776
1
>>> foo()
165989788
165989776
1
>>> foo()
165989788
165989776
1
