Python def function return None in all cases

Question

def leap(year):
    if year % 4 == 0:
        if year % 100 == 0:
            if year % 400 == 0:
                return True
    else:
        return False


print(leap(1992))

It always returns the none value but I want to print the bool in result

Answer 1

Your code

def leap(year):
    if year % 4 == 0:
        if year % 100 == 0:
            if year % 400 == 0:
                return True
    else:
        return False


print(leap(1992))

Let's break it down for 1992.

1. year = 1992.

   1992 % 4 = 0. True. The code moves to next if statement.

   1992 % 100 = 92(False). so your codes halts at this point and exits from the function. Hence the None output for 1992 since you haven't added any logic to return false here.

A simple approach to avoid the None output would be.

def leap(year):
    if year % 4 == 0 and year % 100 == 0 and year % 400 == 0:
        return True
    else:
        return False


print(leap(1992))

Answer 2

Try this:

def leap(year):
    return (year%4==0 and year%100!=0 or year%400==0)

This is equivalent to this with if-else format:

def leap(year):
    if year%4 == 0:
        if year%100 == 0:
            if year%400 == 0:
                return True
            else:
                return False
        else:
            return True
    else:
        return False

You missed out the else parts, and for those cases, the function won't return any boolean value and return None . The above function can be simplified with removing the return False conditions inside else parts and add one at the last of the function.

def leap(year):
    if year%4 == 0:
        if year%100 == 0:
            if year%400 == 0:
                return True
        else:
            return True
    return False