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Python Python Basics (2015) Number Game App Squared

Agustin Fitipaldi
Agustin Fitipaldi
1,644 Points
Posted by Agustin Fitipaldi
Agustin Fitipaldi
1,644 Points

not sure whats wrong here...

tried to do this squared function, but ended up confused.

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

3 Answers

Katie Wood
Katie Wood
19,141 Points
Katie Wood
Katie Wood
19,141 Points

Hi there,

Your code is mostly correct, and will work in most situations. The issue occurs when you feed in a number as a string, like your example ("2"). Let's walk through why. Say we call squared using "2" - we'd get here first:

try:
        int(num)   #this will pass, so it will skip the ValueError and go to the else block

else:
        return num * num    #here's where the issue happens

This will cause an issue because, while it did determine that num could be converted to an integer, the actual value of num is still "2". That means it is still trying to multiply the number as a string.

The fix that is closest to what you've got now would be to change the int(num) line into:

num = int(num)   #would set the value also, so it will work when it tries to multiply

You have some other options, though. Here's one:

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

Hope this helps!

Anibal Marquina
Anibal Marquina
9,523 Points
Anibal Marquina
Anibal Marquina
9,523 Points 
# Check you Try block.
# If the argument (arg) can be turned into an integer then is a number... return the square value of that argument.
# If the argument (arg) cannot be turned into an integer.. then return the argument multiplied by its length.

def squared(arg):
    try:
        int(arg)
        return int(arg) ** 2
    except:
        return arg* len(arg)
Henrik Christensen
seal-mask
.a{fill-rule:evenodd;}techdegree
Henrik Christensen
Python Web Development Techdegree Student 38,322 Points
Henrik Christensen
.a{fill-rule:evenodd;}techdegree
Henrik Christensen
Python Web Development Techdegree Student 38,322 Points

Here is another solution if you prefer to use the else block

def squared(num):
    try:
        num = int(num)  # add 'num =' here :-)
    except ValueError:
        return num * len(num)
    else:
        return num * num