map() method of
TypedArray instances creates a new typed array populated with the results of calling a provided function on every element in the calling typed array. This method has the same algorithm as
map(callbackFn) map(callbackFn, thisArg)
A function to execute for each element in the typed array. Its return value is added as a single element in the new typed array. The function is called with the following arguments:
The current element being processed in the typed array.
The index of the current element being processed in the typed array.
The typed array
map()was called upon.
A value to use as
callbackFn. See iterative methods.
A new typed array with each element being the result of the callback function.
Array.prototype.map() for more details. This method is not generic and can only be called on typed array instances.
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]
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