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Python Python Basics (2015) Number Game App Squared

squared() function code challenge (no "else"?) [Solved]

The squared challenge mentions that you "might" not need the else block...

Here's my code

def squared(A_arg):
   try:
        num = int(A_arg)
   except ValueError:
        return(A_arg * len(A_arg))
   else:
        return(num ** 2)

I just can't see how I could get around not using the else block without an error. Would it makes since (or even work)?

[MOD: added ```python markdown formatting -cf]

Chris Freeman
Chris Freeman
Treehouse Moderator 67,989 Points

Great Job! It looks correct to me. Marking post as Solved.

1 Answer

Ohhhh! I think I figured it out.

def squared(A_arg):
   try:
        num = int(A_arg)
   except ValueError:
        return(A_arg * len(A_arg))
   return(num **2)

I think return ends the fuction. So i just put return(num ** 2) where the else block was and it worked! Sorry for the lack of indentation on my code. It was indented when I posted question.