TypedArray.prototype.sort()

Try it

const uint8 = new Uint8Array([40, 10, 50, 20, 30]);
uint8.sort();

console.log(uint8);
// Expected output: Uint8Array [10, 20, 30, 40, 50]

Syntax

sort()
sort(compareFn)

Parameters

compareFn Optional

A function that determines the order of the elements. The function is called with the following arguments:

a

The first element for comparison.

b

The second element for comparison.

It should return a number where:

• A negative value indicates that a should come before b.
• A positive value indicates that a should come after b.
• Zero or NaN indicates that a and b are considered equal.

To memorize this, remember that (a, b) => a - b sorts numbers in ascending order.

If omitted, the typed array elements are sorted according to numeric value.

Return value

The reference to the original typed array, now sorted. Note that the typed array is sorted in place, and no copy is made.

Description

See Array.prototype.sort() for more details. This method is not generic and can only be called on typed array instances.

Examples

Using sort()

For more examples, see also the Array.prototype.sort() method.

let numbers = new Uint8Array([40, 1, 5, 200]);
numbers.sort();
// Uint8Array [ 1, 5, 40, 200 ]
// Unlike plain Arrays, a compare function is not required
// to sort the numbers numerically.

// Regular Arrays require a compare function to sort numerically:
numbers = [40, 1, 5, 200];
numbers.sort();
// [1, 200, 40, 5]

numbers.sort((a, b) => a - b); // compare numbers
// [ 1, 5, 40, 200 ]

Specifications

Browser compatibility

See also

• Polyfill of TypedArray.prototype.sort with modern behavior like stable sort in core-js
• JavaScript typed arrays guide
• TypedArray
• TypedArray.prototype.reverse()
• TypedArray.prototype.toSorted()
• Array.prototype.sort()