[FIXED] Return of None not expected

Issue

I have a small function as follows:

# Given the numerical value of a minute hand of a clock, return the number of degrees assuming a circular clock.
# Raise a ValueError for values less than zero or greater than 59.


def exercise_20(n):
    try:
        if n < 0 or n > 59:
            raise ValueError("Number supplied is less than 0 or greater than 59")
    except ValueError as ex:
        print(ex)
    else:
        return n * 6


print(exercise_20(-15))
print(exercise_20(30))
print(exercise_20(75))

Here is the output:

Number supplied is less than 0 or greater than 59
None
180
Number supplied is less than 0 or greater than 59
None

Why am I returning 'None' when I strike an exception?

The function correctly prints the exception for the appropriate values and it prints the correct answer for a value within the correct range.

I don't understand why it also printing 'None' when it strikes an exception.

Solution

your function should just raise the error. let the called to the catching:

def exercise_20(n):
    if n < 0 or n > 59:
        raise ValueError("Number supplied is less than 0 or greater than 59")
    return n * 6

for n in (-15, 30, 75, 30):
    print(f"{n=}:", end=" ")
    try:
        print(exercise_20(n))
    except ValueError as ex:
        print(ex)

it will output:

n=-15: Number supplied is less than 0 or greater than 59
n=30: 180
n=75: Number supplied is less than 0 or greater than 59
n=30: 180

you code returns None because you do not have an explicit return statement after print(ex). therefore python implicitly returns None.

Answered By - hiro protagonist