Question

Possible Duplicate:
“Least Astonishment” in Python: The Mutable Default Argument

Consider the following function:

def foo(L = []):
  L.append(1)
  print L

Each time I call foo it will print a new list with more elements than previous time, e.g:

>>> foo()
[1]
>>> foo()
[1, 1]
>>> foo()
[1, 1, 1]

Now consider the following function:

def goo(a = 0):
  a += 1
  print a

When invoking it several times, we get the following picture:

>>> goo()
1
>>> goo()
1
>>> goo()
1

i.e. it does not print a larger value with every call.

What is the reason behind such seemingly inconsistent behavior?
Also, how is it possible to justify the counter-intuitive behavior in the first example, why does the function retain state between calls?

Answer 1

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