A question on modification of a list via a function invocation

[oliver]

It is not a global because accessing an item of a list does not change
whether it is global or local. It would only be global if you declared it
global via a "global" statement. Can you give an example, that would help
determine the issue.

On Mon, 14 Aug 2017 at 16:06 Mok-Kong Shen <mok-kong.shen at t-online.de>
wrote:

> Am 14.08.2017 um 21:53 schrieb Ned Batchelder:
> > On 8/14/17 3:21 PM, Mok-Kong Shen wrote:
> >> Am 14.08.2017 um 20:50 schrieb Ned Batchelder:
> >>> On 8/14/17 2:21 PM, Mok-Kong Shen wrote:
> >>>> I ran the attached program and got the following output:
> >>>>
> >>>> [1, 2, 3]
> >>>> [3, 6, 9]
> >>>>
> >>>> I don't understand why the modification doesn't work in the case of
> >>>> test() but does work in the case of test1().
> >>>>
> >>>> Thanks for your help in advance.
> >>>>
> >>>> M. K. Shen
> >>>>
> >>>> ------------------------------------------------------------
> >>>>
> >>>> def test(alist):
> >>>>     alist=[3,6,9]
> >>>>     return
> >>>>
> >>>> def test1(alist):
> >>>>     alist[0],alist[1],alist[2]=3,6,9
> >>>>     return
> >>>>
> >>>> ss=[1,2,3]
> >>>> test(ss)
> >>>> print(ss)
> >>>> test1(ss)
> >>>> print(ss)
> >>>
> >>> This reassigns the name alist:  alist = [3, 6, 9].   That changes the
> >>> local variable, but cannot affect the caller's variables.
> >>>
> >>> This leaves alist as the same object, but reassigns its elements,
> >>> mutating the list:  alist[0] = 3
> >>>
> >>> This talk has more details: https://nedbatchelder.com/text/names1.html
> >>
> >> I could more or less understand that in test() alist is interpreted as
> >> local but in the extended program below in test2() I first write the
> >> same as in test1(), after which I logically assume that the name alist
> >> is now known as global and then I write alist=[30,60,90] but that
> >> doesn't have any effect globally, since I get the output:
> >>
> >> [1, 2, 3]
> >> [3, 6, 9]
> >> [3, 6, 9]
> >>
> >> Could you please explain that?
> >>
> >> M. K. Shen
> >> ---------------------------------------------------------
> >>
> >> def test(alist):
> >>    alist=[3,6,9]
> >>    return
> >>
> >> def test1(alist):
> >>    alist[0],alist[1],alist[2]=3,6,9
> >>    return
> >>
> >> def test2(alist):
> >>    alist[0],alist[1],alist[2]=3,6,9
> >>    alist=[30,60,90]
> >>    return
> >>
> >> ss=[1,2,3]
> >> test(ss)
> >> print(ss)
> >> test1(ss)
> >> print(ss)
> >> test2(ss)
> >> print(ss)
> >
> > Your test2 function first mutates the caller's list by assigning
> > alist[0]=3, then it rebinds the local name alist to be a new list.  So
> > the caller's list is now [3, 6, 9].
>
> Sorry for my poor knowledge. After the line alist[0]..., what is the
> status of the name alist? It's now a global name, right? So why in the
> line following that the name alist would suddenly be interpreted as
> local? I can't yet fully comprehend the logic behind that.
>
> M. K. Shen
>
>
>   .
>
> --
> https://mail.python.org/mailman/listinfo/python-list
>
-- 
Oliver
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