What's the output?

Question

```javascript
const myPromise = Promise.resolve(Promise.resolve('Promise'));

function funcOne() {
  setTimeout(() => console.log('Timeout 1!'), 0);
  myPromise.then(res => res).then(res => console.log(`${res} 1!`));
  console.log('Last line 1!');
}

async function funcTwo() {
  const res = await myPromise;
  console.log(`${res} 2!`)
  setTimeout(() => console.log('Timeout 2!'), 0);
  console.log('Last line 2!');
}

funcOne();
funcTwo();
```

Accepted Answer

First, we invoke `funcOne`. On the first line of `funcOne`, we call the _asynchronous_ `setTimeout` function, from which the callback is sent to the Web API. (see my article on the event loop here.) Then we call the `myPromise` promise, which is an _asynchronous_ operation. Pay attention, that now only the first then clause was added to the microtask queue. Both the promise and the timeout are asynchronous operations, the function keeps on running while it's busy completing the promise and handling the `setTimeout` callback. This means that `Last line 1!` gets logged first, since this is not an asynchonous operation. Since the callstack is not empty yet, the `setTimeout` function and promise in `funcOne` cannot get added to the callstack yet. In `funcTwo`, the variable `res` gets `Promise` because `Promise.resolve(Promise.resolve('Promise'))` is equivalent to `Promise.resolve('Promise')` since resolving a promise just resolves it's value. The `await` in this line stops the execution of the function until it receives the resolution of the promise and then keeps on running synchronously until completion, so `Promise 2!` and then `Last line 2!` are logged and the `setTimeout` is sent to the Web API. If the first then clause in `funcOne` had its own log statement, it would be printed before `Promise 2!`. Howewer, it executed silently and put the second then clause in microtask queue. So, the second clause will be printed after `Promise 2!`. Then the call stack is empty. Promises are _microtasks_ so they are resolved first when the call stack is empty so `Promise 1!` gets to be logged. Now, since `funcTwo` popped off the call stack, the call stack is empty. The callbacks waiting in the queue (`() => console.log("Timeout 1!")` from `funcOne`, and `() => console.log("Timeout 2!")` from `funcTwo`) get added to the call stack one by one. The first callback logs `Timeout 1!`, and gets popped off the stack. Then, the second callback logs `Timeout 2!`, and gets popped off the stack.

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