Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Python Python Basics (2015) Number Game App Squared

This challenge is similar to an earlier one. Remember, though, I want you to practice! You'll probably want to use try a

Im stuck on this challenge

squared.py
# EXAMPLES
# squared(5) would return 25
# squared("2") would return 4
# squared("tim") would return "timtimtim"
def square(argum): 
    try:
        if argum = int:
            return argum*argum
        except ValueError:
            return len(argum)* argum

3 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,423 Points

You have the right idea. To check is something is an int you can use isinstance(argum, int) to test object type. (Taught in later lesson)

Since you have a try block, you can also remove the if and just "go for it" by using argum = int(argum)

Remember to align the except with the try

Post back if you need more help. Good luck!!!

This is what i am getting now: SyntaxError: argum = int(argum):

that is after i remove the "if", and correct the rest of the line like you said.

It worked!!!!! Thank you!!!!