On 18/07/2008, Christopher Spears <[EMAIL PROTECTED]> wrote:
> How come the stack doesn't shrink when I pop off the last value? I tested
> the code in the interpreter:
>
> >>> lista = [1,2,3,4]
> >>> lista[:len(lista)-1]
> [1, 2, 3]
> >>> lista = lista[:len(lista)-1]
> >>> lista
> [1, 2, 3]
First, a tip:
Instead of lista[:len(lista)-1], you can (and should) just write lista[:-1].
Now, what if we wrap that in a function:
>>> def shorten(lst):
... lst = lst[:-1] # identical to: lst = lst[:len(lst)-1]
...
Then test it:
>>> lista = [1, 2, 3, 4]
>>> shorten(lista)
What do you think will be the result of:
>>> print lista
?
PS. You wrote:
> def stackpop(self):
> length = len(self)
> if length == 0:
> print "Empty list!"
If you're trying to pop from an empty stack, this is an error
condition. Rather than printing a message, the correct way to handle
errors in python is by raising exceptions. e.g.
def stackpop(self):
if len(self) == 0:
raise IndexError # or you could define your own exception...
> def peek(self):
> length = len(self)
> if length == 0:
> return 0
The same applies here -- peeking at an empty stack should be an error
too. Otherwise, how can you tell the difference between an empty
stack and a stack where the top item happens to be a 0?
--
John.
_______________________________________________
Tutor maillist - Tutor@python.org
http://mail.python.org/mailman/listinfo/tutor
