Coding challenge | How to check if a number is odd or even using JS

Determining whether a number is odd or even is a fundamental programming task. In JavaScript, there are several ways to accomplish this. This tutorial will walk you through two simple methods - using the modulus operator and leveraging the bitwise AND operator.

1. Using the modulus operator

The modulus operator (%) is the most common and straightforward way to determine if a number is odd or even. This operator returns the remainder of a division operation.

Key concept:

• A number is even if it is divisible by 2 (remainder = 0).
• A number is odd if it is not divisible by 2 (remainder ≠ 0).

Code example:

function isEven(number) {
    return number % 2 === 0;
}

function isOdd(number) {
    return number % 2 !== 0;
}

// Usage:
console.log(isEven(4)); // true
console.log(isOdd(4));  // false
console.log(isEven(7)); // false
console.log(isOdd(7));  // true

Output:

• isEven(4) returns true, meaning 4 is even.
• isOdd(7) returns true, meaning 7 is odd.

2. Using the bitwise AND operator

The bitwise AND operator (&) can also be used to determine odd or even numbers. This approach relies on binary representation:

Key concept:

• Even numbers have a 0 as the least significant bit (e.g., 2 = 10 in binary).
• Odd numbers have a 1 as the least significant bit (e.g., 3 = 11 in binary).

How it works:
Performing number & 1 checks the last bit of the number:

• If number & 1 === 0, the number is even.
• If number & 1 === 1, the number is odd.

Code example:

function isEven(number) {
    return (number & 1) === 0;
}

function isOdd(number) {
    return (number & 1) === 1;
}

// Usage:
console.log(isEven(4)); // true
console.log(isOdd(4));  // false
console.log(isEven(7)); // false
console.log(isOdd(7));  // true

Output:

• isEven(4) returns true, meaning 4 is even.
• isOdd(7) returns true, meaning 7 is odd.

Comparison of methods:

Feature	Modulus Operator %	Bitwise AND Operator &
Readability	Very easy to understand	Less intuitive
Readability	Slightly slower for large numbers	Slightly faster for large numbers
Use Case	General-purpose applications	Optimised low-level operations

Which method should you use?

• Use the Modulus operator if you want a solution that is simple and easy to understand. It’s the most common method and is suitable for most scenarios.
• Use the bitwise AND operator if performance is critical (e.g., in high-performance code or embedded systems).

Final thoughts

Both methods are effective for checking if a number is odd or even in JavaScript. Choose the one that best fits your specific use case. Happy coding! ?

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