[Tutor] RE: Newbie list question
Tobin, Mark
Mark.Tobin@attcanada.com
Fri, 13 Jul 2001 11:21:59 -0600
Does:
foo += 'c'
act like an append then? I always assumed it was the same as:
foo = foo + 'c'
which obviously should raise a TypeError. Here however it works, in that it
appends 'c' to the object to which foo refers and thus to the object to
which bar refers...
Mark
-----Original Message-----
From: python-list-admin@python.org
[mailto:python-list-admin@python.org]On Behalf Of Joshua Marshall
Sent: Friday, July 13, 2001 12:52 PM
To: python-list@python.org
Subject: Re: Newbie list question
Matthew Alton <Matthew.Alton@anheuser-busch.com> wrote:
...
>>>> foo = ['a', 'b', 'c'] # We have a list named 'foo.' Excellent.
>>>> bar = foo # bar points to foo. Or does it?
bar points to the same object that foo points to. It's not the case
that bar is an alias for foo.
>>>> baz = foo[:] # baz is a copy of foo.
For clarity, it might be better to say baz points to a copy of the
list which foo points to.
>>>> foo
> ['a', 'b', 'c']
>>>> bar
> ['a', 'b', 'c']
>>>> baz
> ['a', 'b', 'c'] # So far, so good.
>>>> del foo[2] # Get rid of 'c' in foo and, therefore in
> bar (?)
>>>> foo
> ['a', 'b'] # 'c' is gone from foo...
>>>> bar
> ['a', 'b'] # ... and also from bar, as expected.
>>>> baz
> ['a', 'b', 'c'] # baz, the copy, is unaffected. Also as
> expected.
>>>> foo = foo + ['c'] # Add 'c' back to foo.
Here, the variable foo is rebound to a new list. The previous list
(which bar still points to) is unaffected. If you had done
"foo.append('c')" instead of "foo = foo + ['c']", than a 'c' would be
appended to the list object that foo and bar both still refer to.
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