Creating a (ES6) promise without starting to resolve it

I want to add my 2 cents here. Considering exactly the question "Creating a es6 Promise without starting resolve it" I solved it creating a wrapper function and calling the wrapper function instead. Code:

Let's say we have a function f which returns a Promise

/** @return Promise<any> */
function f(args) {
   return new Promise(....)
}

// calling f()
f('hello', 42).then((response) => { ... })

---

Now, I want to prepare a call to f('hello', 42) without actually solving it:

const task = () => f('hello', 42) // not calling it actually

// later
task().then((response) => { ... })

Hope this will help someone :)

---

Referencing Promise.all() as asked in the comments (and answered by @Joe Frambach), if I want to prepare a call to f1('super') & f2('rainbow'), 2 functions that return promises

const f1 = args => new Promise( ... )
const f2 = args => new Promise( ... )

const tasks = [
  () => f1('super'),
  () => f2('rainbow')
]

// later
Promise.all(tasks.map(t => t()))
  .then(resolvedValues => { ... })

Good question!

The resolver passed to the promise constructor intentionally runs synchronous in order to support this use case:

var deferreds = [];
var p = new Promise(function(resolve, reject){
    deferreds.push({resolve: resolve, reject: reject});
});

Then, at some later point in time:

 deferreds[0].resolve("Hello"); // resolve the promise with "Hello"

The reason the promise constructor is given is that:

• Typically (but not always) resolution logic is bound to the creation.
• The promise constructor is throw safe and converts exceptions to rejections.

Sometimes it doesn't fit and for that it the resolver runs synchronously. Here is related reading on the topic.