Why does this aysnc function not implicitly return a promise?

Question

I read in this Stack Overflow question async/await implicitly returns promise? that:

"An Async function will always return a promise. If you don't explicitly return a promise, the value you return will automatically be wrapped in a promise."

So why does my getData2() function below return undefined?

const getData = async() => {
  return new Promise(resolve => {
    setTimeout(() => {
      resolve('this is the data');
    }, 2000);
  });
};

const getData2 = async() => {
  setTimeout(() => {
    return 'this is the data without an explicit promise';
  }, 1000);
};

(async() => {
  console.log(await getData2()); // DOES NOT WORK: displays 'undefined' immediately
  console.log(await getData()); // WORKS: waits one second and displays data
})();

You've got a function without a `return` statement. It *does* return a Promise, but the value is `undefined` because the function returns `undefined`. — Pointy, Sep 02 '21 at 02:04
Look at what function your `return` statement is inside of. In `getData2()`, the `return is returning from the `setTimeout()` callback function, NOT from your `getData2()` function. There is no `return` statement in your actual `getData2()` function so it returns `undefined` which means the promise that `getData2()` returns, resolves with `undefined`. — jfriend00, Sep 02 '21 at 02:11
It does return a promise but the promise gets resolved immediately since you don't await anything inside the function. And since you don't return anything from the function (all it does is call `setTimeout()`) it resolves with `undefined`. — Lennholm, Sep 02 '21 at 02:20

score 3 · Accepted Answer · answered Sep 02 '21 at 02:03

getData2 is returning a Promise, but you're awaiting it, which extracts the resolve value from the Promise.

console.log(await getData2());

turns into

console.log(await promiseThatResolvesToUndefined);

console.log(undefined);

If you don't await it, you'll see the Promise.

const getData2 = async () => {
    setTimeout(() => {
        return 'this is the data without an explicit promise';
    }, 1000);
};

(async () => {
    console.log(getData2());
})();

The resolve value is undefined because nothing is being returned inside the body of getData2.

score 1 · Answer 2 · answered Sep 02 '21 at 02:05

The issue is not with async but with using return inside an inner function (in this case a lambda), which returns from that inner function (not the outer function), and setTimeout then ignores that return value. The correct way to do this is getData (but without the async keyword which adds an second layer of Promise).

Why does this aysnc function not implicitly return a promise?

2 Answers2