Default values for function args

[RootBoy]

On Tue, 11 Jan 2000 04:05:49 GMT, Joshua Macy <amused at webamused.com> wrote:
 >> 
 >>   Maybe you're thinking that assignment will create a copy of an object,
 >> instead of a copy of a reference to an object?  In Python assignment
 >> always stores references to objects, not copies of the objects
 >> themselves.  In a statement like
 >> 
 >> target = expression
 >> 
 >> expression yields an object, and the assignment stores a reference to
 >> the object in target.  If the expression is simply the name of a
 >> variable, after the assignment both variables have references to the
 >> same object.  This is what's happening in the list example you created.
 >>   To create a copy of an object, you have to take an extra step, as in
 >> the example I gave where I used the slice operation ([:]) to create a
 >> copy of the sequence that contained everything the original sequence
 >> did, and then assigned the reference to that new object to my variable. 
 >> For instance, if explicitly created a copy of local_list before
 >> returning the reference, your function would have behaved as you
 >> expected:
 >> 
 >> 
 >> >>> def fn(a, local_list=[]):   # local_list object gets created here
 >> ... 	local_list.append(a)
 >> ... 	return local_list[:]    # a copy of local_list is created and
 >> returned
 >> ... 
 >> >>> list1 = fn('xyzzy')
 >> >>> print list1
 >> ['xyzzy']
 >> >>> list2 = fn(3.14)
 >> >>> print list1, list2
 >> ['xyzzy'] ['xyzzy', 3.14]
 >> >>> list3 = fn(1000)
 >> >>> print list1, list2, list3
 >> ['xyzzy'] ['xyzzy', 3.14] ['xyzzy', 3.14, 1000]
 >> >>> 
 >> 
 >>   Does that help at all?

Yes, it's finally clear.  I had overlooked an important footnote in the
tutorial:

   
"Actually, call by object reference would be a better description, since 
if a mutable object is passed, the caller will see any changes the callee
makes to it (e.g., items inserted into a list). "

Although this refers to passing arguments into a function, it also apparently
applies to return values.  I had thought I was returning an entire list object
by passing the entire list by value on the stack.  What actually happens is
that a reference to the object is what gets passed back to the caller by value
on the stack.

Thanks again....