Others have already explained why it works the way it does, but
haven't given an example of how to get a default argument value of a
new empty list each time the function is called. 


On Sat, Aug 11, 2001 at 02:42:50PM -0100, Simon Vandemoortele wrote:

| --- quote ---
| Important warning: The default value is evaluated only once. This makes a 
| difference when the default is a mutable object such as a list or dictionary. 
| For example, the following function accumulates the arguments passed to it on 
| subsequent calls: 
| 
| def f(a, l = []):
|     l.append(a)
|     return l
| print f(1)
| print f(2)
| print f(3)
| 
| This will print 
| 
| [1]
| [1, 2]
| [1, 2, 3]
| --- end quote ---

Try 

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


'None' is an immutable object.  This means it can't change and will
always be the same.  The first thing I do in the function is to see if
'l' has the default value.  If it does I create a new local binding to
a new empty list object called 'l'.  Then I proceed to modify it and
return it as before.  This idiom will have the desired effect.

HTH,
-D


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