A TypeError prevents me from forming the conditionals of my function (Python)

Question

The function is supposed to receive a number representing a year, and then print if it's a leap year or not.

def isItALeapYear(year):

    while True: 
        if year % 4 == 0:
            print("That is a leap year! ")
            break
    
        elif year % 4 != 0:
            print("That is not a leap year...")
            break

        elif not isinstance(year, int) or year == None:
            print("Please enter a number...")
            break

The program works, the only thing I can't get right is that it is supposed to notify you if anything that it's not a number is being used as an argument. I've tried both the isinstance() function, as well as writing what I want as year != int. And then year == None in the hopes of making it work in case anything nondefined is used as an argument.

I read this post with the exact same error: TypeError: not all arguments converted during string formatting python

But I'm not intending to format anything with the % symbol As far as I'm concerned the % can be used as an operand to get the residue of a division. So in this case I'm using it to figure out if a year is a leap year or not by asking if the residue is 0 when divided by 4. I'm pretty stuck, and the sad thing is the error comes up in the very first "if", so I can't really know if the last lines for excluding any non int type argument work or not. Any help would be really appreciated!

Answer 1

I recommend to divide the functionality of checking the number from returning the output as well as from receiving the input.

def is_multiple_of_four(number: int):
    if number % 4 == 0:
        return True
    else:
        return False


if __name__ == '__main__':
    user_input = ""
    while not user_input.isdigit():
        user_input = input("Please type in a year: ")

    if is_multiple_of_four(int(user_input)):
        print("That is a leap year!")
    else:
        print("That is not a leap year.")

Here you can see the function that checks the number does only that, it checks the number if it's modulo of 4 equals 0. In the script outside the function you can retrieve the user input for as long as it takes to get a valid numeric and return the output in respect of the functions results.

Edit (adding clarification asked in the comments)

The first condition if __name__ == '__main__' is quiet common in python. It's not necessary for your function, but I like using it in answers if people seem to learn Python, so they don't miss out on it. Here is a question with a good answer: What does if __name__ == "__main__": do? The short answer in the accepted answer is enough to understand why you might want to use it.

The second condition

    user_input = ""
    while not user_input.isdigit():

first defines a variable user_input with an arbitrary non-digit String value and than uses the negated isdigit() method of the String class on it as condition. Therefor the while loop gets entered in the beginning, as the arbitrary value is not an digit. From then on the value will be re-assigned with user input until it holds an actual digit. This digit is still a String however.

Answer 2

You could use the isleap() function from the standard library (module calendar):

from calendar import isleap

def isItALeapYear(year):
    if not isinstance(year, int):
        print("Please provide a number")
    elif isleap(year):
        print("That is a leap year!")
    else:
        print("That is not a leap year...")

Answer 3

First thing while loop in the function makes no sense you can remove it(If you want).

There are multiple ways to do that I will show.

First One.

def isItALeapYear(year):
    if type(year) != int:
        return         # here return to exit the function 
    while True: 
        if year % 4 == 0:
            print("That is a leap year! ")
            break
    
        elif year % 4 != 0:
            print("That is not a leap year...")
            break

        elif not isinstance(year, int) or year == None:
            print("Please enter a number...")
            break

Another is.

def isItALeapYear(year):
    try:
        int(year)
    except ValueError: # this line of code executes when the year is not the int
        return  # here return to exit the function 
    while True: 
        if year % 4 == 0:
            print("That is a leap year! ")
            break
    
        elif year % 4 != 0:
            print("That is not a leap year...")
            break

        elif not isinstance(year, int) or year == None:
            print("Please enter a number...")
            break

I know there are more ways to do that but these are the best ones (I Think).

Your function is not accurate then you can use this one.

def isItALeapYear(year):
    if type(year) != int:
        return
    if (( year%400 == 0)or (( year%4 == 0 ) and ( year%100 != 0))):
        print(f"{year} is a Leap Year")
    else:
        print(f"{year} is Not the Leap Year")

Edit For Quetioner

def isItALeapYear(year):
    if type(year) != int:
        return
    if (( year%400 == 0)or (( year%4 == 0 ) and ( year%100 != 0))):
        print(f"{year} is a Leap Year")
    else:
        print(f"{year} is Not the Leap Year")
try:
    isItALeapYear(asdasd)
except NameError:
    print("You give the wrong Value")