Issue Promise.all()

Question

I want to declare an array of empty promises
then resolve them manually, but it is not working ?

const tab = [1]
let promises = tab.map(e => new Promise(() => { }))
setTimeout(() => {
  promises[0] = Promise.resolve()
}, 3000);
Promise.all(promises)
  .then(e => console.log("finished"))

I DO NOT WANT to do it like this :

const tab = [1]
let promises = tab.map(e => new Promise((resolve) => { 
  setTimeout(() => {
    resolve()
  }, 3000);
}))

Promise.all(promises)
  .then(e => console.log("finished"))

What's the use case behind this, can help in understanding what you want to exactly achieve? — qasimalbaqali, Oct 10 '21 at 10:55
The resolve doesn’t resolve with anything? Also I think if you’re using `Promise.resolve` the promise constructor is not needed. — evolutionxbox, Oct 10 '21 at 11:02

score 2 · Accepted Answer · answered Oct 10 '21 at 10:58

2

Thats the closest you can get:

const tab = [1]
let resolves = []
let promises = tab.map(e => new Promise((resolve) => {
   resolves.push(resolve)
}))
setTimeout(() => {
  resolves[0]()
}, 3000);
Promise.all(promises).then(e => console.log("finished"))

answered Oct 10 '21 at 10:58

Rani Sharim

610
3
7

I’m still confused, each individual promise resolves undefined? – evolutionxbox Oct 10 '21 at 11:06
You nailed it – anesboz Oct 10 '21 at 11:10
4

This pattern is bad practice. – trincot Oct 10 '21 at 11:10

score 0 · Answer 2 · answered Oct 11 '21 at 03:03

Here's a technique using a reusable deferred function that allows you to externally control a promise. We extend both resolve and reject functionality, allowing better control for the caller. This is a somewhat unconventional pattern but there are advantages to exploiting promises in this way. See this Q&A for an detailed example -

function deferred () {
  let resolve, reject, promise = new Promise((res, rej) => {
    resolve = res; reject = rej
  })
  return { promise, resolve, reject }
}

const tasks = [ deferred(), deferred(), deferred() ] // 3 "deferrred" objects

Promise.all(tasks.map(t => t.promise)).then(console.log, console.error)

setTimeout(_ => tasks[0].resolve("one"), 1000)
setTimeout(_ => tasks[1].resolve("two"), 2000)
setTimeout(_ => tasks[2].resolve("three"), 3000)

Issue Promise.all()

2 Answers2