Are async functions really functions or Promise objects?

Question

I was just trying different combinations of async await problem statements and I tried this,
basically I have two functions promise1() and promise2() which return two promises which resolve values 10 and 20 after 3 and 1 seconds respectively. Below is the code that I have written

function promise1()
{
        return new Promise((resolve,reject)=>{
                setTimeout(()=>{
                        resolve(10)
                },3000)
        })
}

function promise2()
{
        return new Promise((resolve,reject)=>{
                setTimeout(()=>{
                        resolve(20)
                },1000)
        })
}

async function sum()
{
        let num1 = await promise1();
        let num2 = await promise2();

        return num1+num2;
}
sum().then(s => console.log(s))

when I execute this code I am getting 30 as answer

as in the above code if I was able to apply .then() to sum then is it a promise?
when I did console.log(typeof sum) its saying sum is a function not an object so what exactly are async functions? I have tried searching the answer for this behaviour but couldn't find any sources that would correctly answer. It would be a great help for me if someone answered this query or tell me the online sources or books that would get me the answer
my node version is v10.15.3

thank you

`sum` is a function that returns a Promise. `sum()` calls that function and evaluates to a Promise, just like `const fn = () => 3`, `fn` is a function, `fn()` evaluates to a number — CertainPerformance, Apr 27 '19 at 08:34
Depending on your code editor, hovering with the mouse over a variable will tell you exactly what its type is! — Kokodoko, Apr 27 '19 at 08:36

score 3 · Answer 1 · answered Apr 27 '19 at 08:34

They are functions where the return value is always a promise.

async function foo() { }

const value = foo();

console.log(foo instanceof Function);
console.log(foo instanceof Promise);
console.log(value instanceof Function);
console.log(value instanceof Promise);

as in the above code if I was able to apply .then() to sum then is it a promise?

No, you weren't.

sum.then() won't work. sum().then() will. Adding () will call a function and give you the return value.

T.J. Crowder · Accepted Answer · 2019-04-27T08:42:38.533

Are async functions really functions or Promise objects?

They're really functions. The functions return promises.

as in the above code if I was able to apply .then() to sum then is it a promise?

You didn't. You used .then() on the result of calling sum (sum().then(...), not sum.then(...)). It's just like calling a function that returns a string, then using toUpperCase on what the function returns.

These two functions are essentially equivalent, handwaving some minor details:

// #1
function foo() {
    return new Promise((resolve, reject) => {
        try {
            getSomePromise()
            .then(result => result * 2)
            .catch(reject);
        } catch (error) {
            reject(error);
        }
    });
}

and

// #2
async function foo() {
    const result = await getSomePromise();
    return result * 2;
}

Naturally, that's not how you'd probably write #1 in real life, it's an example of the promise creation antipattern, but it's a reasonable interpretation of what an async function looks like under the hood. Purely for completeness, you'd probably write #1 like this:

// #3
function foo() {
    return getSomePromise().then(result => result * 2);
}

The difference is that if getSomePromise throws an error rather than running and returning a promise, that function (#3) will throw an error (synchronously), whereas #1 above will return a promise that rejects instead. async functions do the latter, which is why #1 is written the way it is above.

Are async functions really functions or Promise objects?

2 Answers2