Async function without await in Javascript

Question

I have two functions a and b that are asynchronous, the former without await and the later with await. They both log some

Answer 1

Everything is synchronous until a Javascript asynchronous function is executed. In using async-await await is asynchronous and everything after await is placed in event queue. Similar to .then().

To better explain take this example:

function main() {
  return new Promise( resolve => {
    console.log(3);
    resolve(4);
    console.log(5);
  });
}

async function f(){
    console.log(2);
    let r = await main();
    console.log(r);
}

console.log(1);
f();
console.log(6);

As await is asynchronous and rest all is synchronous including promise thus output is

1
2
3
5
6
// Async happened, await for main()
4

Similar behavior of main() is without promise too:

function main() {
    console.log(3);
    return 4;
}

async function f(){
    console.log(2);
    let r = await main();
    console.log(r);
}

console.log(1);
f();
console.log(5);

Output:

1
2
3
5
// Asynchronous happened, await for main()
4

Just removing await will make whole async function synchronous which it is.

function main() {
    console.log(3);
    return 4;
}

async function f(){
    console.log(2);
    let r = main();
    console.log(r);
}

console.log(1);
f();
console.log(5);

Output:

1
2
3
4
5