The JavaScript array.sort() method

Q: Why does sort() put 10 before 2?

Because the default sort compares elements as strings, and "10" sorts before "2". Pass a compare function — (a, b) => a - b — to sort numerically.

Q: How do I sort numbers in JavaScript?

Use a compare function: arr.sort((a, b) => a - b) for ascending, (a, b) => b - a) for descending.

Q: How do I sort an array of objects?

Compare the property you care about: users.sort((a, b) => a.age - b.age) for numbers, or a.name.localeCompare(b.name) for text.

Method JavaScript All modern browsers Updated June 23, 2026

Quick answer

The sort() method orders an array in place and returns it. By default it sorts elements as strings, which is why [10, 2, 1].sort() surprisingly gives [1, 10, 2]. For numbers, pass a compare function: arr.sort((a, b) => a - b) sorts ascending. It mutates the original array.

Overview

sort() rearranges an array into order and returns that same (now sorted) array. The headline gotcha is the default: with no arguments it converts every element to a string and sorts those, so [10, 2, 1] becomes [1, 10, 2] — because "10" comes before "2" alphabetically. For text that's fine; for numbers it's almost never what you want.

The fix is a compare function. You give sort() a function that takes two elements, a and b, and returns a negative number if a should come first, a positive number if b should, or 0 if they tie. For numbers ascending, (a, b) => a - b is the whole trick; flip it to b - a for descending. For objects, compare a property: (a, b) => a.age - b.age.

Two things to keep in mind: sort() mutates the array, so copy it first with slice() or the spread operator if you need to keep the original; and modern engines make it stable, so equal elements keep their relative order. To reverse an already-sorted array, there's reverse().

Syntax

array.sort()                 // sorts as strings (watch out for numbers)
array.sort(compareFn)

array.sort((a, b) => a - b)  // numbers ascending
array.sort((a, b) => b - a)  // numbers descending

Parameters

The array.sort() method accepts the following parameters.

Parameter	Description
`compareFn`	Optional. Function comparing two elements `(a, b)`. Return a negative number to put `a` first, positive to put `b` first, `0` to keep order. Omit it and elements are sorted as strings.

Example

Live example

<pre id="out" style="font:15px ui-monospace,monospace"></pre>
<script>
  const nums = [10, 2, 33, 4, 1];

  const wrong = [...nums].sort();           // string sort
  const right = [...nums].sort((a, b) => a - b);

  document.getElementById('out').textContent =
    'default: ' + wrong.join(', ') + '\n' +
    'numeric: ' + right.join(', ');
  // default: 1, 10, 2, 33, 4 / numeric: 1, 2, 4, 10, 33
</script>

Best practices

Always pass a compare function for numbers — (a, b) => a - b — or the default string sort will mis-order them.
Copy the array first ([...arr].sort(...)) if you need to keep the original, since sort() mutates.
Sort objects by a key with (a, b) => a.value - b.value, or localeCompare for strings.
For descending order, swap the operands: (a, b) => b - a.

Frequently asked questions

Why does sort() put 10 before 2?

Because the default sort compares elements as strings, and "10" sorts before "2". Pass a compare function — (a, b) => a - b — to sort numerically.

How do I sort numbers in JavaScript?

Use a compare function: arr.sort((a, b) => a - b) for ascending, (a, b) => b - a) for descending.

Does sort() change the original array?

Yes, it sorts in place and returns the same array. Copy it first with [...arr] or slice() if you need the original.

How do I sort an array of objects?

Compare the property you care about: users.sort((a, b) => a.age - b.age) for numbers, or a.name.localeCompare(b.name) for text.

array.some() array.splice()

References