Once-only evaluation of default parameter values function definitions

[Fred Ma]

Michael Geary wrote:

> >   Example#1
> >   ---------
> >   def f(a, L=[]):
> >       L.append(a)
> >       return L
> >
> >   Example#2
> >   ---------
> >   def f(a, L=None):
> >       if L is None:
> >           L = []
> >       L.append(a)
> >       return L
>
> Here's a simpler example:
> 
> >>> a = []
> >>> b = a
> >>> a
> []
> >>> b
> []
> >>> a.append(1)
> >>> a
> [1]
> >>> b
> [1]
> 
> See what happened? The statement 'b = a' didn't make a copy of a and
> store it in b. Instead, it made the name b refer to the same object
> as a. So, when I said b.append(1), it meant exactly the same thing
> as if I'd said a.append(1).

Thanks, Mike.  That's alot clearer.  The default value looks like it's
a persistent object.  It is similar to a static local variable in C++.
In Example#1, that object actually got changed.  In contrast, in
Example#2, the code prevents L from remaining bound to None beyond the
first line, so the None object never gets changed.  In fact, my 2nd
explanation was close: that there is an unnamed persistent object that
holds the default value to be bound to L for invocations that don't
explicity supply an argument for L. The place I went wrong was to say
that L never gets bound to None again, after it has been bound to a
caller supplied object on a previous invocation of the function.

Fred