[Tutor] Functions returning multiple values

[Giorgio]

Thankyou Hugo!

Ok, so i think the key is of my problem is that when doing X = 0 i'm
creating a new object, that only exist in the local namespace. BUT, when
using a list as a parameter for a function i'm only giving it a new name,
but the object it's referring to it's always the same, and is in the global
namespace.

Right?

2010/2/23 Hugo Arts <hugo.yoshi at gmail.com>

> On Tue, Feb 23, 2010 at 1:13 PM, Giorgio <anothernetfellow at gmail.com>
> wrote:
> > I have an update:
> > I can easily undertand why this example doesn't work:
> > def nochange(x):
> >     x = 0
> > y = 1
> > nochange(y)
> > print y # Prints out 1
> > X is a local variable, and only gets modified in the function, that
> doesn't
> > return any value.
> > But it's very difficult for me to understand WHY this works:
> > def change(some_list):
> >     some_list[1] = 4
> > x = [1,2,3]
> > change(x)
> > print x # Prints out [1,4,3]
> > some_list is a "local" list, isn't it? Maybe i can't have lists that are
> > only existing in a function?
>
> Here is what happens, as I understand it:
> When you enter the function, a new name (x, in your case) is created
> in the local scope. That name points to the object you supplied when
> you called the function (an integer object, with a value of 1). the x
> = 0 statement creates a new object, and has the name x now pointing to
> this new object. The old integer object still exists, and y still
> points to it. This is why the global y name is not affected by the
> change in x
>
> Now, in the second example, the same thing basically happens. A new
> name is created and pointed at the object you supplied. However, the
> statement some_list[1] = 4 is different from the assignment, in that
> it doesn't create a new object; It modifies the existing one. Since
> the global and local names both point to the same object, the change
> you make is reflected in both.
>
> You can of course create a new list object, so that the original is
> not affected:
>
> def nochange(some_list):
>    # create shallow copy of list. NOTE: Shallow copies may still bite
> you if you change the list members.
>    some_list = some_list[:]
>    some_list[1] = 2
>
> >>> x = [1, 2, 3]
> >>> nochange(x)
> >>> x
> [1, 2, 3]
>
> HTH,
> Hugo
>



-- 
--
AnotherNetFellow
Email: anothernetfellow at gmail.com
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