Code execution not happening in asyn function

Question

In the following code the last line does not print ('end'):

function setTimeoutSync(fn, milli) {
    return new Promise((resolve) => {
        setTimeout(fn, milli);
    });
}

async function run () {
    console.log('start');
    await setTimeoutSync(() => { console.log('setTimeoutSync'); }, 3000);
    console.log('end');
}

run();

It is not that the script is exiting, because nothing I put after the await statement gets executed.

How can NodeJS just not execute statements in a function? Also, what's a good way to do a good old-fashioned synchronous wait in NodeJS? This question is actually more about the former than the latter.

When I call `resolve` it prints 'end' before 'setTimeoutSync'. — HenryTK, Jul 19 '19 at 09:42
See the linked question on how to convert a callback API to promises — CertainPerformance, Jul 19 '19 at 09:44
You should not pass a callback to a function that returns a promise — Bergi, Jul 19 '19 at 09:52
there's nothing sync about `setTimeoutSync`, even if you call it like that. Better write a simple promise-wrapper for setTimeout `const wait = delay => new Promise(resolve => setTimeout(resolve, delay));` and use it as `await wait(3000); console.log('setTimeoutSync'); console.log('end');` — Thomas, Jul 19 '19 at 09:55

Sagiv b.g · Accepted Answer · 2019-07-19T10:00:23.787

Not sure if this is a typo but you forgot to call resolve so the next line after the await will execute:

function setTimeoutSync(fn, milli) {
  return new Promise((resolve) => {
    setTimeout(() => {
      fn()
      resolve()
    }, milli);
  });
}

async function run() {
  console.log('start');
  await setTimeoutSync(() => {
    console.log('setTimeoutSync');
  }, 3000);
  console.log('end');
}

run();

As for your question:

Can you explain how not calling resolve can result in the rest of the function not executing?...

Well async await uses generators and promise behavior:

The purpose of async/await functions is to simplify the behavior of using promises synchronously and to perform some behavior on a group of Promises. Just as Promises are similar to structured callbacks, async/await is similar to combining generators and promises.

A generator can yield results thus not terminating the execution context, and can continue it where it left off.

Basically your example could be written with Promise.prototype.then() callback:

function setTimeoutSync(fn, milli) {
  return new Promise((resolve) => {
    setTimeout(() => {
      fn()
      resolve()
    }, milli);
  });
}

async function run() {
  console.log('start');
  setTimeoutSync(() => {
    console.log('setTimeoutSync');
  }, 3000)
  .then((result) => {
   console.log('end');
  });
  
}

run();

So as you see, if we are not resolving, the callback to .then won't run.

Thanks. Can you explain how not calling `resolve` can result in the rest of the function not executing? Is it causing the `run` function to return early? I would expect a runtime error if statements in a function were not executable. — HenryTK, Jul 19 '19 at 09:48
It was not knowing that `generators` existed that confused me. Now it makes sense. Thanks! — HenryTK, Jul 19 '19 at 10:08

Code execution not happening in asyn function

1 Answers1