Question

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

I recently met a problem in Python.
Code:

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

 print f(1)
 print f(2)
 print f(3)

the output would be

[1]
[1, 2]
[1, 2, 3]

but why the value in the local list: L in the function f remains unchanged?
Because L is a local variable, I think the output should be:

[1]
[2]
[3]

I tried another way to implement this function:
Code:

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

This time, the output is:

[1]
[2]
[3]

I just don’t understand why…
Does anyone have some ideas? many thanks.

Answer 1

The default parameters are in fact initialized when the function is defined, so

def f(L = []): pass

is quite similar to

global default_L = []
def f(L = default_L): pass

You can see this way that it is the same list object that is used in every invocation of the function.