JavaScript Program to Check for a Leap Year

A leap year is a year that is divisible by 4, but if it is also divisible by 100, it should be divisible by 400 to be a leap year.

In this tutorial, we will write JavaScript programs to check if a given year is a leap year using both simple logic and functions.

1. Basic Leap Year Check Using if…else Statements

Here’s a straightforward JavaScript program to check if a year is a leap year.

// Program to check if a year is a leap year
function isLeapYear(year) {
    if ((year % 4 === 0 && year % 100 !== 0) || (year % 400 === 0)) {
        return true;
    } else {
        return false;
    }
}

// Test the function
const year = 2024;
if (isLeapYear(year)) {
    console.log(`${year} is a leap year.`);
} else {
    console.log(`${year} is not a leap year.`);
}

Output:

2024 is a leap year

Explanation of the Code

Define the isLeapYear Function:

function isLeapYear(year) {
    if ((year % 4 === 0 && year % 100 !== 0) || (year % 400 === 0)) {
        return true;
    } else {
        return false;
    }
}

This function takes a year as an argument.
It uses an if…else statement to check if the year is a leap year using the following conditions:
year % 4 === 0: Checks if the year is divisible by 4.
year % 100 !== 0: Checks if the year is not divisible by 100, which covers the rule that every 100th year is not a leap year unless it is also divisible by 400.
year % 400 === 0: Checks if the year is divisible by 400, making it a leap year.

If either of the conditions is satisfied, the function returns true, indicating it is a leap year.
If none of the conditions are satisfied, it returns false.

Test the Function:

const year = 2024;
if (isLeapYear(year)) {
    console.log(`${year} is a leap year.`);
} else {
    console.log(`${year} is not a leap year.`);
}

Calls the isLeapYear function with the year (in this case, 2024).

Uses console.log() to display the result based on the function's return value.

2. Shortened Leap Year Check Using a Ternary Operator

You can also write a more concise version of the leap year check using the ternary operator.

function isLeapYear(year) {
    return (year % 4 === 0 && year % 100 !== 0) || (year % 400 === 0);
}

// Test the function
const year = 1900;
console.log(`${year} is ${isLeapYear(year) ? 'a leap year' : 'not a leap year'}.`);

Output:

1900 is not a leap year.

Explanation of the Shortened Code

Simplified isLeapYear Function:

This version uses the same condition but directly returns the boolean result of the expression:

return (year % 4 === 0 && year % 100 !== 0) || (year % 400 === 0);

There’s no need for an if…else statement here, making the code shorter and cleaner.

Using the Ternary Operator:

console.log(`${year} is ${isLeapYear(year) ? 'a leap year' : 'not a leap year'}.`);

Uses the ternary operator (? 🙂 to decide the output based on the return value of isLeapYear(year).

3. Inline Leap Year Check Without a Function

If you don't need to reuse the leap year check and want to keep it inline, here's a quick way to perform the check:

const year = 2000;
const isLeap = (year % 4 === 0 && year % 100 !== 0) || (year % 400 === 0);

console.log(`${year} is ${isLeap ? 'a leap year' : 'not a leap year'}.`);

Output:

2000 is a leap year.

Explanation

Directly stores the result of the leap year check in the variable isLeap.
Uses the same condition as before to determine if year is a leap year.
The result is then logged using console.log().

Summary

The logic to check for a leap year in JavaScript follows these steps:

A year is a leap year if it is divisible by 4.
However, if the year is also divisible by 100, it must be divisible by 400 to be a leap year.

This tutorial shows you multiple ways to implement a leap year check in JavaScript:

Using an if…else statement within a function.
Using a concise version with a ternary operator.
Inline check without defining a separate function.

You can use any of these approaches depending on your use case and coding style preferences.