async/await always returns promise

Question

I'm trying async/await functionality. I have such code imitating a request:

const getJSON = async () => {
  const request = () => new Promise((resolve, reject) => (
    setTimeout(() => resolve({ foo: 'bar'}), 2000)
  ));

  const json = await request();
  return json;
}

When I use the code in this way

console.log(getJSON()); // returns Promise

it returns a Promise

but when I call this line of code

getJSON().then(json => console.log(json)); // prints { foo: 'bar' }

it prints json as expected

Is it possible to use just code like console.log(getJSON())? What don't I understand?

Answer 1

Every async function returns a Promise object. The await statement operates on a Promise, waiting until the Promise resolves or rejects.

So no, you can't do console.log on the result of an async function directly, even if you use await. Using await will make your function wait and then return a Promise which resolves immediately, but it won't unwrap the Promise for you. You still need to unwrap the Promise returned by the async function, either using await or using .then().

When you use .then() instead of console.logging directly, the .then() method makes the result of the Promise available to you. But you can't get the result of the Promise from outside the Promise. That's part of the model of working with Promises.

Answer 2

Return value of an async function will always be an AsyncFunction Object, which will return a Promise when called. You can not change that return type. The point of async/await is to easily wait for other async process to complete inside an async function.

Answer 3

A function defined with async always returns a Promise. If you return any other value that is not a Promise, it will be implicitly wrapped in a Promise. The statement const json = await request(); unwraps the Promise returned by request() to a plain object { foo: 'bar' }. This is then wrapped in a Promise before being returned from getJSON so a Promise is what you ultimately get when you call getJSON(). So to unwrap it, you can either call getJSON().then() like you've done or do await getJSON() to get the resolved value.

Answer 4

const getResOrErr = () => {
  const callAsyncCodeHere = async () => {
  const request = () =>
      new Promise((resolve, reject) =>
        setTimeout(() => resolve({ foo: "bar" }), 2000)
      );

    const json = await request();
    return json;
  };
  return callAsyncCodeHere()
   .then(console.log)
   .catch(console.log);
};

getResOrErr();

Try this. You can achieve making a function inside your main function and then put you promise code inside that function. Call it there and when you get the response or error just return it.