Use async await with

Given the following code:

var arr = [1,2,3,4,5];

var results: number[] = await (item): Promise<number> => {
        await callAsynchronousOperation(item);
        return item + 1;

which produces the following error:

TS2322: Type 'Promise<number>[]' is not assignable to type 'number[]'. Type 'Promise<number> is not assignable to type 'number'.

How can I fix it? How can I make async await and work together?

The problem here is that you are trying to await an array of promises rather than a promise. This doesn't do what you expect.

When the object passed to await is not a Promise, await simply returns the value as-is immediately instead of trying to resolve it. So since you passed await an array (of Promise objects) here instead of a Promise, the value returned by await is simply that array, which is of type Promise<number>[].

What you need to do here is call Promise.all on the array returned by map in order to convert it to a single Promise before awaiting it.

According to the MDN docs for Promise.all:

The Promise.all(iterable) method returns a promise that resolves when all of the promises in the iterable argument have resolved, or rejects with the reason of the first passed promise that rejects.

So in your case:

var arr = [1, 2, 3, 4, 5];

var results: number[] = await Promise.all( (item): Promise<number> => {
    await callAsynchronousOperation(item);
    return item + 1;

This will resolve the specific error you are encountering here.

There's another solution for it if you are not using native Promises but Bluebird.

You could also try using, mixing the and Promise.all

In you case:

  var arr = [1,2,3,4,5];

  var results: number[] = await, async (item): Promise<number> => {
    await callAsynchronousOperation(item);
    return item + 1;

If you map to an array of Promises, you can then resolve them all to an array of numbers. See Promise.all.

I'd recommend using Promise.all as mentioned above, but if you really feel like avoiding that approach, you can do a for or any other loop:

const arr = [1,2,3,4,5];
let resultingArr = [];
for (let i in arr){
  await callAsynchronousOperation(i);
  resultingArr.push(i + 1)

Solution below to process all elements of an array asynchronously AND preserve the order:

const arr = [1, 2, 3, 4, 5, 6, 7, 8];
const randomDelay = () => new Promise(resolve => setTimeout(resolve, Math.random() * 1000));

const calc = async n => {
  await randomDelay();
  return n * 2;

const asyncFunc = async () => {
  const unresolvedPromises = => calc(n));
  const results = await Promise.all(unresolvedPromises);


Also codepen.

Notice we only "await" for Promise.all. We call calc without "await" multiple times, and we collect an array of unresolved promises right away. Then Promise.all waits for resolution of all of them and returns an array with the resolved values in order.