Squared Code Challenge

Spencer:

The problem is in your try::except block. If the try block succeeds, nothing is returned because your return statement is indented under your exception case. Also, you should store the result in a variable and return the result. You could do something like this.

try:
    result = int(num ** 2)
except ValueError:
    print("please enter a valid number.")
return result

Yes. The reason you are getting a TypeError when a string is passed is because the int() function cannot convert a string representing characters (i.e.; "tim"). To handle that you need to setup your except case to catch TypeError. So for my example provided previously, the except part would look something like.

except TypeError:
      result = num * len(num)

I ended up assigning num to an int before trying to do math against it. Like so:

def squared(num):
    try:
        num = int(num)
        result = num * num

TL;DR - Change TypeError back to ValueError

So, I had to take the code and execute it on the command line to play with everything when I was working through the issue. If you modify your code to accept some input and pass it through to your function, you will start to see the errors.

Anyway, your code will pass if you change TypeError back to ValueError. I don't fully understand the difference between the two, but when I ran the code with exception TypeError it failed and told me I had a ValueError instead.

You could also solve it this way:

def squared(num):
    try:
        result = num * num
    except TypeError:
       result = num * len(num)
    return result

The reason you are getting a ValueError is because you are trying to convert a string to int() and it is failing. If you take away the int() part, then you need to check for TypeError in the except clause. To test this type the following into a Python interpreter:

>>> num = 'hello'
>>> num * num

results in TypeError

>>> num = int(num)

results in a ValueError.