Thanks, I though it was a reference (tough to implement I'm sure) Regards,
Philippe Peter Hansen wrote: > Philippe C. Martin wrote: >> l = ['ABCDE','FGHI'] > > Okay so far... > >> l[1:] #returns ['FGHI'] > > Which is a _copy_ (via slicing) of part of the list. Another way of > saying this is that it is a _new_ list which has a copy of the > references from the appropriate part of the old list. > > Try "l[1:] is l[1:]" to prove that... > >> l[1:][0] #return 'FGHI' > > Sure does. From the new list. > >> a = l[1:][0] + 'J' #a becomes 'FGHIJ' > > Because you are actually storing a reference to the new list, whose > first element you have modified. > >> l[1:][0] += 'J' #NO ERROR BUT l[1:][0] == 'FGHI' > > You are modifying the first element of the *copy* of the slice of the > list, but you don't ever store a copy of it. When you try to check what > happened with the second part, you are creating yet another copy of part > of the list and sure enough the original has never been changed. > >> What am I missing ? > > That slicing makes copies. If you directly access the element in the > first list (without using a slice) it will work. > > (I think I've got most of the correct...) > > -Peter -- http://mail.python.org/mailman/listinfo/python-list
