Why `async/await` doesn't work in my case?

Question

I read about async/await, but I've a critical question. At first I explain an old example to show base of my question and then I ask my exact question.

Everybody know it:

console.log('1');
console.log('2');
console.log('3'); // Ex: 123

It is simple but in below case:

console.log('1');
setTimeout(()=>{
    console.log('2');
},0);
console.log('3'); // Ex: 132

It is simple too, setTimeout function is asynchronous and JavaScript jump from it and after resolve run its function, so we see 2 after 1 and 3.

But, now I read async/await and I wrote a function like this:

(async function test() {
    console.log('1');
    await setTimeout(()=>{
        console.log('2');
    },0);
    console.log('3');
})(); // Ex: 132

The Export is 132 too, why? this is my question, why 3 run before 2? I expect because of async/await after 1 JavaScript wait for 2 and then wrote 3. why 132?

Jollanta

async functions are part of ES2017, not ES7 (ES2016). – Mongolism

@FelixKling, thanks, I'm confusing, is ES2017 called ES8? – Jollanta

Answer 1

await only suspends when the value passed to it is a Promise. In your case, setTimeout returns a Number so await doesn't wait for it.

The correct code will be as follows:

async function test() {
    console.log('1');
    await new Promise((resolve, reject) => {
        setTimeout(() => {
            console.log('2');
            resolve()
        }, 0);
    });
    console.log('3');
}

Curate

Answer 2

Because setTimeout doesn't return a promise. await x only waits if x is a promise; if x isn't a promise, it's (effectively) wrapped in one as though you had await Promise.resolve(x). That means the code following it will run asynchronously but as soon as possible.*

If you want a promise version of setTimeout, see this question's answers. But even with that, your test function wouldn't use a callback, instead you'd just await the promise-enabled timeout:

function later(delay) {
    return new Promise(function(resolve) {
        setTimeout(resolve, delay);
    });
}

async function test() {
  console.log("1");
  await later(10);
  console.log("2");
  console.log("3");
}

test().catch(e => console.error(e));
console.log("After the call (just to prove it does wait for the timeout after 1 and before 2");

---

* On browsers, that's guaranteed to be before a setTimeout(..., 0) scheduled during the same task, because promise callbacks scheduled during a task occur just after the end of that task, before the next task is picked up from the queue (even if the next task was scheduled before the promise callback). More on this ("macrotasks" and "microtasks") in this question's answers.

Hardback

Answer 3

You can await to that functions that returns Promise. setTimeout does not returns Promise. So in this case await used before the setTimeout does not has a sense.

You can wrap your setTimeout into a Promise and call resolve at the setTimeout function.

(async function test() {
    console.log('1');
    await new Promise((resolve, reject) => { 
         setTimeout(() => {
            console.log('2');
            resolve(); // also can pass a parameter here to get it via await.
         },0);
    });
    console.log('3');
})();

Elman