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Python Python Basics (2015) Number Game App Squared

Posted by Alexander Davison
Alexander Davison
65,469 Points

Squared function

It says "Bummer! 'squared' didn't return the correct result"

What's wrong???

squared.py 
# EXAMPLES
# squared(5) would return 25
# squared("2") would return 4
# squared("tim") would return "timtimtim"

def squared(arg):
  try:
    if int(arg) == arg:
      return arg ** 2
  except:
    try:
      return int(arg) ** 2
    except:
      return arg * len(arg)

1 Answer

Jennifer Nordell
seal-mask
STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Ok you're checking in your code to see if the int version of the argument is equal to the argument. If it is, return the square of it. Then if it isn't possible to make it an int... you check it again to see if it can be made an int and returned as a square directly. When that fails then you send back the string printed out a certain number of times. But you don't say what kind of exception you're looking to "catch".

Somewhere in all that you managed to overthink it a bit, I think. Here's the code I used. I hope it helps!

def squared(arg):
  try:
    return int(arg) ** 2
  except ValueError:
    return arg * len(arg)
Alexander Davison
Alexander Davison
65,469 Points

Thanks you so mush! :D