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Python Python Basics (2015) Number Game App Squared

Posted by moe sal
moe sal
920 Points

Not sure what the error "squared" didn't return the right answer" implies. Help!

it seems like I have done everything the description asks for.

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

3 Answers

Gabbie Metheny
Gabbie Metheny
33,778 Points
Gabbie Metheny
Gabbie Metheny
33,778 Points

The problem is your if statement: you don't need one! Try basically takes care of that part for you. Python will try to convert your argument into an integer, and if it can't, then the except will run. So, remove the if statement, and make sure you're setting your argument to equal the integer version of the argument, and it should pass!

Wesley Trayer
Wesley Trayer
13,812 Points
Wesley Trayer
Wesley Trayer
13,812 Points

Hey, Moe! The problem lies in the line:

int(argument)

The integer form of "argument" isn't stored in a variable. Your code works for numbers, or a string of letters, but it fails when it receives a string filled with numbers.

Hope this helps! :)

Oszkár Fehér
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.a{fill-rule:evenodd;}techdegree seal-36
Oszkár Fehér
Treehouse Project Reviewer
Oszkár Fehér
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Oszkár Fehér
Treehouse Project Reviewer

Hi Moe Your code it's ok just it doesn't catch one more case

# squared("2") would return 4

a string which is an integer for that it should be another if statement

 if argument == str(int(argument)):
     return int(argument) ** 2

Also the first return method, returns argument multiplied with himself instead of square

return argument ** 2

Happy coding