Python how to check if it is a leap year with ==int

 def leapyear(year):
if year/400 == int :
    return False
if year/100 == int :
    return False
if year/4 == int :
    return True

hello I would like to know why my code doesn't work with it == to int because essentially its the same thing as using modulo and == to 0 this is just a question that came to me.

def check(n): if n > 200: return "large" 
x = n/2 if x !=int: return "odd" 
elif x==0 and n< 100: return "small" 
elif x==0 and n>100: return "medium"

also how come the int works here

Solution

Your issue is that int is a type. If you try to compare a number to a type of object, which is what you are doing when you write if year/400 == int :, it will always return False, because these can never be the same.

A better way to check if year/400 is an integer would be:

if year%400 == 0:
    return False

This is saying: If the remainder of year/400 is equal to 0, return False, which is what you wanted.

Some other things are:

You should use elif instead of if in most cases. In this one, it doesn't matter, since the return statement terminates the execution of the function early, but you should still use it anyways. Otherwise, even when you have your final result, it will go through the rest of the if statements.

The other thing isn't related to your code, but your leap year criteria are incorrect. If the year is divisible by 4, then it's a leap year. Unless the year is divisible by 100, then it's not. Unless it's divisible by 400, then it is.

Improved code:

def leapyear(year):
    if year%400 == 0:
        return True
    elif year%100 == 0:
        return False
    elif year%4 == 0:
        return True
    return False