Is_prime returns None


#1



https://www.codecademy.com/courses/python-intermediate-en-rCQKw/0/6?curriculum_id=4f89dab3d788890003000096#

I've been working on is_prime for a while now, and I don't understand what is wrong with my code. It tells me that is_prime(2) returns None instead of False.

I made special cases for all x below 2, and used a while loop to check if x divided by some number n is zero or not. There are return statements and everything. What is structurally wrong with this code that makes it not return True or False?


def is_prime(x):
    if x <= 1:
        return False
    else:
        n = 2
        while n < x:
            if x%n==0:
                return False
                break
            else:
                n += 1
                return True


#2

None represents lack of value, a function that doesn't explicitly return something will return None to say that it had no return value.

Can you think of a way that could happen for your function?
You could start by verifying that None is the result of is_prime(2) as the error message claims.


#3

I fixed it by adding another special case for 2, and, interestingly enough, changing the while loop to a for loop. shrugs Go figure.


#4

Yes. Do! There's something to learn there.

And, a very important thing to notice in your code is that your while loop always returns on the first iteration, it isn't a loop at all.


#5

def is_prime(x):
    if x<2:
        return False
    elif x == 2:
        return True
    for n in range(2, x):
        if x%n == 0:
            return False
    return True

The last for loop is unnecessary, because the code will loop through all of the numbers between 2 and x. That means you only need to return true only if x can't evenly divide the numbers in between.
e.g: if x is 19, and we start at 2, so is 2 evenly divisible by 19? 3? 4?... so if all of the numbers from 2-18 can't be divided evenly, we will return True.


#6

:relaxed:great answer