Quick asynchronous in js

Arnav Zek
1 min readApr 25, 2019


By promises

  • Define a promise:
  • establish a method: promise.then(…..after promise is completed)

let lazy = new Promise(resolve=>setTimeout(()=>resolve(‘I am lazy’) ,1000))
lazy.then(result => console.log(result) )

in place of setTimeout there can be a matrix calculation or a network request

without arrow function parent’s ‘this’ context won’t be passed


By async and await

  • function should be marked as an async function
  • define your lazy function function as a promise and mark the line as await

async function asyncCall() {
for(i=1; i<=10; i++){
await new Promise(resolve=>setTimeout(()=>resolve(console.log(i)) ,100))



Arnav Zek

Freelancer . Fulstack Dev . Game Dev