DEV Community: Saharat Paynok

I've been drifting, here is my line in the sand.

Saharat Paynok — Wed, 24 Sep 2025 00:55:05 +0000

I've been distracting myself instead of committing.
I've been busy but not productive.
I've been wasting me potential.

Here's my line in the sand:

I will stop letting Youtube and distractions steal my focus
I will publish progress every week, no matter how small.
I will ship one AI side project live in the next 90 days
I will hold my self accountable by being public about this journey

This post is my first step in burning the old identity

If you’re reading this — hold me accountable.
Call me out if I disappear.

New chapter starts today.

Do not use array.sort() Function Without a Compare Function in JavaScript

Saharat Paynok — Sat, 18 May 2024 15:01:22 +0000

You might need to sort their elements when working with arrays in JavaScript. The sort() function is commonly used for this purpose. However, there’s an important caveat: if you’re dealing with an array of numbers, using sort() without a compare function can yield unexpected outcomes.

The Default Behavior of sort()

By default, the sort() function treats elements as strings and compares them based on their UTF-16 code units order. This behavior works well for sorting strings alphabetically, but it falls short when sorting numbers

Example 1: Sorting Strings

Let’s say we have an array of animal names:

var animals = ["Horse", "Cat", "Tiger", "Lion"];
animals.sort(); // Result: ["Cat", "Horse", "Lion", "Tiger"]

The strings are sorted alphabetically, which is what we expect.

Example 2: Sorting Numbers

Now, consider an array of numbers:

var numbers = [12, 10, 15, 11, 14, 13, 16];
numbers.sort(); // [10, 11, 12, 13, 14, 15, 16]

Wait, what happened? The numbers are sorted as strings! The reason is that the default comparison treats them as such.

The Solution: Using a Compare Function

You must provide a compare function to sort numbers (or any other data type) properly. This callback function defines an alternative sorting order. It takes two arguments (often referred to as a and b) and returns either a positive, negative, or zero.

Here’s how you can create a simple compare function for sorting numbers:

function compare(a, b) {
    return a - b;
}

In this function:

If a is greater than b, it returns a positive value.
If a equals b, it returns zero.
If a is less than b, it returns a negative value.
Now, let’s sort our array of numbers using this compare function:

numbers.sort(compare); // Result: [10, 11, 12, 13, 14, 15, 16]

Perfect! The numbers are now sorted correctly.

Conclusion

Remember that when sorting arrays in JavaScript, especially with numbers, always use a compare function. It ensures predictable and accurate sorting behavior. Don’t rely on the default string-based sorting; take control by providing your comparison logic.

Happy coding! 🚀

Feel free to customize this article further or add any additional insights you’d like to share. If you have more requests or questions, feel free to ask! 😊👍

How to check if the character is alphanumeric in GO

Saharat Paynok — Thu, 16 May 2024 17:50:11 +0000

Understanding Characters in Go as Bytes

In Go, a string is equivalent to a slice of bytes.

Solution: Checking if a Character is Alphanumeric

To determine whether a character is alphanumeric (i.e., a letter or a digit), you can create a function that checks the byte value of the character.

Here’s an example implementation:

package main

func isAlphaNumeric(c byte) bool {
    // Check if the byte value falls within the range of alphanumeric characters
    return (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') || (c >= '0' && c <= '9')
}

func main() {

    isAlphaNumeric('a') // true

    isAlphaNumeric('2') // true (digits are also considered alphanumeric)

    isAlphaNumeric('#') // false
}

In the isAlphaNumeric function:
We check if the byte value c corresponds to a lowercase letter ('a' to 'z'), an uppercase letter ('A' to 'Z'), or a digit ('0' to '9').
If the condition is met, we return true; otherwise, we return false.

Let me know if you have an alternative solution or learned something from this post.