[Tutor] Python Function Doubt

[Dave Angel]

nikhil <nik.mis at gmail.com> wrote:

> Hi,
>
> I am learning Python and came across this example in the Python Online
> Tutorial (
> http://docs.python.org/tutorial/controlflow.html#default-argument-values)
> for Default Argument Values.
>
> ex:
>
> def  func(a, L=[ ]):
> L.append(a)
> return L
>
> print func(1)
> print func(2)
> print func(3)
>
> *O/P*
> [1]
> [1,2]
> [1,2,3]
>
> Now my doubt is,
>
> 1)  After the first function call, when ' L' (Reference to List object) goes
> out of scope, why isn't it  destroyed ?
>
> 2) It seems that this behavior is similar to static variable logic in C
> language.  Is it something similar ? OR
>     Is it related to scope rules in Python ?
>
>
>   
The L parameter is a default parameter.  That's a formal parameter with 
an initial value (L=[]).  The reference for that value is established 
when the definition is executed, not when the function is called.  The 
net effect when the function is at top-level is similar to a static 
variable in C.  But it's not really L that is retained, but the object 
its bound to.  Perhaps it's easiest to think that the function object 
has a reference to the same object, and copies it to L each time the 
function is called without that particular formal parameter.

See the online help  (this one's for 2.6.2, but other recent versions 
are the same):
   
http://www.python.org/doc/2.6.2/reference/compound_stmts.html#function-definitions

Usually, you want to make such an initial value a non-mutable value, 
such as None, 42, or "".  But if it's mutable, it will indeed retain its 
value from one call to the next of that function.  As you observed.

There are times when this behavior is useful (eg. nested definitions), 
and I think it's permanently established in the lanaguage.  But it is 
perhaps one of the most common beginner confusions.  And it's part of 
the solution to some of the other common problems.