Asked  7 Months ago    Answers:  5   Viewed   249 times

I'm trying to understand how async/await works in conjunction together with promises.


async function latestTime() {
  const bl = await web3.eth.getBlock('latest');
  console.log(bl.timestamp); // Returns a primitive
  console.log(typeof bl.timestamp.then == 'function'); //Returns false - not a promise
  return bl.timestamp;
const time = latestTime(); // Promise { <pending> }


As far as I understand, await should be blocking and in the code above it seemingly blocks returning an object bl with the primitive timestamp. Then, my function returns the primitive value, however the time variable is set to a pending promise instead of that primitive. What am I missing?



An async function always returns a promise. That's how it reports the completion of its asynchronous work. If you're using it in another async function, you can use await to wait for its promise to settle, but in a non-async function (often at the top level or in an event handler), you have to use the promise directly, e.g.:

.then(time => {
.catch(error => {
    // Handle/report error

If you're doing this at the top level of a JavaScript module, some environments now support the upcoming top-level await in modules:

const time = await latestTime();

JavaScript engines are getting support for top-level await, and Webpack has experimental support for it, for instance.

Here's a rough translation of your async function in explicit Promise terms:

function latestTime() {
    return new Promise((resolve, reject) => {
        .then(bl => {
            console.log(typeof bl.timestamp.then == 'function');

Some important notes on that:

  • The function you pass to new Promise (the promise executor function) gets called synchronously by new Promise.
    • Which is why the operation starts, web3.eth.getBlock is called synchronously to start the work.
  • Any error (etc.) thrown within the promise executor gets caught by new Promise and converting into a promise rejection.
  • Any error (etc.) thrown within a promise callback (like the one we're passing then) will get caught and converted into a rejection.
Tuesday, June 1, 2021
answered 7 Months ago

The promise will always log pending as long as its results are not resolved yet. You must call .then on the promise to capture the results regardless of the promise state (resolved or still pending):

let AuthUser = function(data) {
  return google.login(data.username, data.password).then(token => { return token } )

let userToken = AuthUser(data)
console.log(userToken) // Promise { <pending> }

userToken.then(function(result) {
   console.log(result) // "Some User token"

Why is that?

Promises are forward direction only; You can only resolve them once. The resolved value of a Promise is passed to its .then or .catch methods.


According to the Promises/A+ spec:

The promise resolution procedure is an abstract operation taking as input a promise and a value, which we denote as [[Resolve]](promise, x). If x is a thenable, it attempts to make promise adopt the state of x, under the assumption that x behaves at least somewhat like a promise. Otherwise, it fulfills promise with the value x.

This treatment of thenables allows promise implementations to interoperate, as long as they expose a Promises/A+-compliant then method. It also allows Promises/A+ implementations to “assimilate” nonconformant implementations with reasonable then methods.

This spec is a little hard to parse, so let's break it down. The rule is:

If the function in the .then handler returns a value, then the Promise resolves with that value. If the handler returns another Promise, then the original Promise resolves with the resolved value of the chained Promise. The next .then handler will always contain the resolved value of the chained promise returned in the preceding .then.

The way it actually works is described below in more detail:

1. The return of the .then function will be the resolved value of the promise.

function initPromise() {
  return new Promise(function(res, rej) {

  .then(function(result) {
    console.log(result); // "initResolve"
    return "normalReturn";
  .then(function(result) {
    console.log(result); // "normalReturn"

2. If the .then function returns a Promise, then the resolved value of that chained promise is passed to the following .then.

function initPromise() {
  return new Promise(function(res, rej) {

  .then(function(result) {
    console.log(result); // "initResolve"
    return new Promise(function(resolve, reject) {
       setTimeout(function() {
       }, 1000)
  .then(function(result) {
    console.log(result); // "secondPromise"
Tuesday, June 1, 2021
answered 7 Months ago

It's almost like this is being treated as yield await prom. What is going on here?

Exactly that is how async generators behave.

how can I simply yield a rejected promise as-is from this generator.

You cannot. Notice that an async iterator is expected to be consumed by

try {
    for await (const value of orderProms(promises)) {
} catch(err) {
    console.error('Caught error: ', err);

There is no facilitation for individual error handling in the syntax. When there's an exception, the loop stops, the generator is done. Point.

So what can you do? I see three choices:

  • just keep it as is and treat failing early as a feature (similar to Promise.all)
  • handle errors (either in orderProms or before passing promises into it) and yield tuples of promise status and value

    for await (const value of orderProms( =>
        prom.catch(err => `Caught error: ${err}`)
    ))) {
  • use a normal (non-async) generator from which you yield one promise after the other manually, to be able to use it in the way you want
Wednesday, August 18, 2021
answered 4 Months ago


async function addFiles(dir,tree) {
  const files = await readDir(dir)
  await Promise.all( (name) => {await readDir(dir); return name;})
Wednesday, September 15, 2021
answered 3 Months ago

If you have promises that occasionally don't resolve or reject and that's not the way they are supposed to work (which it usually isn't), then you just have to fix that. There really is no work-around. The proper fix is to get down to the lowest level and fix the code so it reliably resolves or rejects every time.

This is not the proper fix, but implementing a timeout wrapper could help with debugging giving you a log message with some semblance of a stack trace for a timed out promise:

function rejectT(t) {
    // create potential error here for better opportunity at stack trace
    let e = new Error("Promise timed out");
    return new Promise((resolve, reject) => {
        setTimeout(() => {
        }, t);

function timeout(p, t = 5000) {
    return Promise.race([p, rejectT(t)]);

You can then wrap any promise such that instead of:


You can use:


Or, if you want to set a custom timeout value:

timeout(fn(), 1000).then(...).catch(...);

Again, this is debugging code to help find the culprits that need fixing and to help test fixes, not to bandaid your code.

Rewriting all the async/await calls to use Promise.then is my last resort.

I don't see how this is going to help at all. If await never finishes, neither will promise.then(). They are exactly the same in that regard. If the promise never resolves or rejects, then the .then() handler will never get called either.

Problem is the code base is riddled with async/await and Promises that sometimes resolve (depending on conditions)

There's no shortcut here other than methodical code review to find suspect code that has code paths that may never resolve or reject and then building unit tests to test every function that returns a promise in a variety of conditions.

One likely source of code that never resolves or rejects are some of the promise anti-patterns. The precise reason some of them are anti-patterns is because they can be very easy to mess up. Here are a few references that might spike your sensitivity to suspect code:

Promise Anti-Patterns

Common Promise Anti-Patterns and How to Avoid Them

ES6 Promises: Patterns and Anti-Patterns

Thursday, November 11, 2021
answered 3 Weeks ago
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