Newbie list question

Mark Day mday at apple.com
Fri Jul 13 19:14:28 CEST 2001


In article <13c69f60.0107130843.5cd7dc38 at posting.google.com>, Matthew
Alton <Matthew.Alton at Anheuser-Busch.COM> wrote:

> >>> foo = ['a', 'b', 'c']  #  We have a list named 'foo.'  Excellent.
> >>> bar = foo              #  bar points to foo.  Or does it?
> >>> baz = foo[:]           #  baz is a copy of foo.
> >>> 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.
> >>> foo
> ['a', 'b', 'c']            #  'c' is back.  Good.
> >>> bar
> ['a', 'b']                 #  ??? What the... ???  Where is 'c'?
> >>> baz
> ['a', 'b', 'c']            #  baz still unaffected, of course.
> >>> 

The statement:
   foo = foo + ['c']
creates a brand new list from the contents of foo and the list ['c']
and assigns the new list to foo.

If you had instead done:
   foo.append('c')
then it would have changed the list that foo and bar were both pointing
at.

-Mark



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