JavaScript Dev Learning Journey - Leap Year!
Sam
Posted on January 3, 2022
I am currently picking back up on assignments I have been working on to improve my JavaScript self-learning journey and currently doing a program to determine if a year given is a Leap Year or not! What better way to start a new year :-)
My first step right away was looking up the method of how to determine if a year is a leap year, which I found out through a Microsoft docs which states the following:
Any year that is evenly divisible by 4 is a leap year: for example, 1988, 1992, and 1996 are leap years.
Which seems pretty easy ! Right away I knew I would be coding a conditional statement so I wrote up the following:
const leapYears = function (yearNum) {
if (yearNum % 4 == 0) {
return true;
}else{
return false;
}
};
Of course, as I was going through the tests, thinking I am in the clear, I knew a part of me felt that this was too easy/good to be true. Sure enough, I ran into a problem where my test failed when the year was 1900 for example. So what I decided to do was go back to documentation to see if there's anything about that and sure enough:
However, there is still a small error that must be accounted for. To eliminate this error, the Gregorian calendar stipulates that a year that is evenly divisible by 100 (for example, 1900) is a leap year only if it is also evenly divisible by 400.
So that means although 1900 works with 100, it doesn't work with 400 and therefore is not a leap year!
So, now I need to account for how to make a number go through several checks:
- Make sure it is divisible by 4.
- Check to see if it's divisible by 100 AND 400 after checking if divisible by 4.
- If it's divisible by 100 but NOT 400, it's not a leap year. Same if it's flipped around.
That's a simple fix! Taking my previous code, I've updated it to the following:
const leapYears = function (yearNum) {
if (yearNum % 4 == 0 && yearNum % 100 == 0 && yearNum % 400 == 0) {
return true;
}else{
return false;
}
//Any year divisible by 4 is a leap year.
//Check if divisible by 3 numbers: 4, 100, and 400.
//if-statements on passing all three conditions to be leap years
};
Where yearNum is checked to have a remainder 0(hence the usage of % to show if it's divisible by a number) with 4, 100, AND 400. This will cover the cases where it can be divisible by 4 and 100 BUT not by 400, making it NOT a leap year. This goes the same if divisible by 100 and 400, but not 4 in addition to divisible by 400 and 4, but not 100.
And no-- 2022 is not a leap year. 2024 is when we will have our next leap year!
Posted on January 3, 2022
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