[Novice] Default argument value in function definition

[Larry Bates]

Default values that are mutable types are evaluated
at definition time not execution time.  That means
L is set to empty list when Python parses function
and it lives outside the function definition.  It
is rather 'different' but I can assure you that is
how it works and it can sometimes bite programmers
new to Python that overlook the docs you refer to.

HTH,
Larry Bates
Syscon, Inc.


"Tito" <titogarcia_nospamniguarreria_ at inicia.es> wrote in message
news:2loejlFelu9nU1 at uni-berlin.de...
> From the Python's tutorial, about default argument values:
>
> <quote>
> The default value is evaluated only once. This makes a difference when the
> default is a mutable object such as a list, dictionary, or instances of
most
> classes. 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]
>
> If you don't want the default to be shared between subsequent calls, you
can
> write the function like this instead:
> def f(a, L=None):
>     if L is None:
>         L = []
>     L.append(a)
>     return L
> </quote>
>
> I can't imagine how subsequent calls to f can share the same value for L.
> The tutorial says that a new symbol table for the variables inside of the
> function is created each time the function is called, and I would say the
> symbol table is destructed when the function finishes execution.
>
> How is the value of L conserved between funtion calls?
> Can someone explain the mechanism to me?
>
> Thanks,
> Tito
>
>