Return a Promise That Resolves After a Delay in JavaScript

The challenge

Write slowAdd(a, b, ms) that returns a promise which resolves to a + b after ms milliseconds.

Write your answer below, then press Check my answer. Your code runs right here in your browser, and the tests tell you what works.

Your solution

It prints

SlowAdd(2, 3, 50) resolves to 5: slowAdd(2, 3, 50) gives 5
It really waits: (async () => { const start = Date.now(); await slowAdd(1, 1, 120); return Date.now() - start >= 100; })() gives true

Stuck? Open one hint at a time. Try again after each one.

Hint 1

return new Promise((resolve) => setTimeout(() => resolve(…), ms));

There’s always more than one right way. Read it, make sure you understand it, then try writing it yourself without looking.

Show one possible solution

Solution

function slowAdd(a, b, ms) {
  return new Promise((resolve) => {
    setTimeout(() => resolve(a + b), ms);
  });
}

This exercise uses what Freshman Year: Asynchronous JavaScript: Waiting Without Freezing teaches. Read those notes first if any of it is new.