Sorting looks like the simplest job an array can do, and then sort() hands you [10,2,33] for an input of [10,2,33] and calls it sorted. The reason is that sort() compares items as strings unless you tell it otherwise. This guide covers how it works, how to sort numbers, strings and objects properly, and the habits that keep it from surprising you.
 

Table of Contents

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What sort() Does

Reorders the array: it arranges the items in a new order.

Changes the original array: it sorts in place, so the array you called it on is modified.

Also returns the array: the same array, now sorted, not a copy.

Sorts as strings by default: every item is converted to text and compared alphabetically.

let fruits = ["mango", "apple", "banana"];
fruits.sort();

console.log(fruits); // ["apple", "banana", "mango"]

Syntax and the Compare Function

array.sort(compareFunction);

compareFunction: optional, but you'll want it for anything other than plain text.

It receives two items, a and b: it decides which one comes first.

Negative result: a goes before b.

Positive result: b goes before a.

Zero: their order stays the same.
 

Sorting Numbers Correctly

The default way goes wrong: numbers get treated as text, so 10 comes before 2.

let nums = [10, 2, 33, 4];
console.log(nums.sort()); // [10, 2, 33, 4], wrong

Ascending order: subtract b from a.

let nums = [10, 2, 33, 4];
nums.sort((a, b) => a - b);
console.log(nums); // [2, 4, 10, 33]

Descending order: subtract a from b.

nums.sort((a, b) => b - a);
console.log(nums); // [33, 10, 4, 2]

 

Sorting Strings

Plain sort works for simple cases: but uppercase letters come before lowercase ones.

let names = ["vi", "Sam", "alex", "Bob"];
console.log(names.sort()); // ["Bob", "Sam", "alex", "vi"]

Use localeCompare for proper alphabetical order:

names.sort((a, b) => a.localeCompare(b));
console.log(names); // ["alex", "Bob", "Sam", "vi"]

It also handles accents and other languages better: which the default sort does not.
 

Sorting Objects

Sort by a number property:

let users = [
  { name: "Vi", age: 22 },
  { name: "Sam", age: 30 },
  { name: "Alex", age: 25 }
];

users.sort((a, b) => a.age - b.age);
console.log(users.map(u => u.name)); // ["Vi", "Alex", "Sam"]

Sort by a string property:

users.sort((a, b) => a.name.localeCompare(b.name));
console.log(users.map(u => u.name)); // ["Alex", "Sam", "Vi"]

 

Things to Remember

It changes the original array: copy it first if you need to keep the original order.

let original = [3, 1, 2];
let sorted = [...original].sort((a, b) => a - b);

console.log(original); // [3, 1, 2]
console.log(sorted);   // [1, 2, 3]

There's a non-mutating version: toSorted() returns a new sorted array and leaves the original alone, in newer JavaScript environments.

let sorted2 = original.toSorted((a, b) => a - b);

Undefined values go to the end: and they never reach your compare function.

Sorting is stable in modern JavaScript: items that compare as equal keep their original relative order.

 

Common Mistakes

Sorting numbers without a compare function: this gives text-based order.

Writing a compare function that returns true or false: it must return a number, negative, zero or positive.

// Wrong
nums.sort((a, b) => a > b);

// Right
nums.sort((a, b) => a - b);

Forgetting that it changes the original: copy the array first when the original order matters.
 

Conclusion

Frequently Asked Questions (FAQs)

Because it compares items as strings by default, and "10" comes before "2" alphabetically.

2.Does sort() change the original array?

Yes. Use [...array].sort() or toSorted() to avoid that.

3.How do I sort in descending order?

Use (a, b) => b - a for numbers, or b.localeCompare(a) for strings.