callbackFn

A callback function that produces an element of the new typed array.

The function is called with the following arguments:

currentValue

The current element being processed in the typed array.

index

The index of the current element being processed in the typed array.

array

The typed array map() was called upon.

thisArg Optional

Value to use as this when executing callbackFn.

A new typed array.

The map() method calls a provided callback function (callbackFn) once for each element in a typed array, in order, and constructs a new typed array from the results.

callbackFn is invoked only for indexes of the typed array which have assigned values; it is not invoked for indexes that are undefined, those which have been deleted, or which have never been assigned values.

callbackFn is invoked with three arguments: the value of the element, the index of the element, and the typed array object being traversed.

If a thisArg parameter is provided to map(), it will be passed to callbackFn when invoked, for use as its this value. Otherwise, the value undefined will be passed for use as its this value. The this value ultimately observable by callbackFn is determined according to the usual rules for determining the this seen by a function.

map() does not mutate the typed array on which it is called (although callbackFn, if invoked, may do so).

The range of elements processed by map() is set before the first invocation of callbackFn. Elements which are appended to the array after the call to map() begins will not be visited by callbackFn. If existing elements of the typed array are changed, or deleted, their value as passed to callbackFn will be the value at the time map() visits them; elements that are deleted are not visited.

The following code takes a typed array and creates a new typed array containing the square roots of the numbers in the first typed array.

const numbers = new Uint8Array([1, 4, 9]);
const roots = numbers.map(Math.sqrt);
// roots is now: Uint8Array [1, 2, 3],
// numbers is still Uint8Array [1, 4, 9]

The following code shows how map() works when a function requiring one argument is used with it. The argument will automatically be assigned to each element of the typed array as map() loops through the original typed array.

const numbers = new Uint8Array([1, 4, 9]);
const doubles = numbers.map((num) => num * 2);
// doubles is now Uint8Array [2, 8, 18]
// numbers is still Uint8Array [1, 4, 9]

• Polyfill of TypedArray.prototype.map in core-js
• TypedArray.prototype.filter()
• Array.prototype.map()