Program terminates after await statement in javascript

Question

I am learning async javascript, and I am struck with a doubt here. Here is my simple code:

function delay(sec)
{
    return new Promise(()=>{
        setTimeout(()=>{
            console.log("Inside settimeout");
        },sec);
    })
}

async function print()
{
    await delay(2000);
    console.log("Outside settimeout");
}

print();

In the above program I expected that due to await keyword, the delay function will wait for 2 seconds, output "Inside settimeout" and then output "Outside settimeout". But instead, it just logs "Inside settimeout" and simply terminates without logging "Outside settimeout". Can someone clear my doubt and explain how this works? Thanks in advance.

You need to call the resolve() function inside your delay function. — Kostas Minaidis
– Kostas Minaidis, Commented Sep 19, 2022 at 8:56
Does this answer your question? Combination of async function + await + setTimeout — Ivar
– Ivar, Commented Sep 19, 2022 at 9:00

Igor Janković · Accepted Answer · 2022-09-19 08:58:26Z

1

You are missing the resolve call in your promise.

function delay(sec)
{
    return new Promise((resolve, reject)=>{
        setTimeout(()=>{
            console.log("Inside settimeout");
            resolve();
        },sec);
    })
}

async function print()
{
    await delay(2000);
    console.log("Outside settimeout");
}

print();

answered Sep 19, 2022 at 8:58

Igor Janković

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2 Comments

tdcoder Over a year ago

I got it...But why did it terminate without logging "Outside settimeout"

Danny Over a year ago

@tdcoder Because the "await" waits for the Promise to resolve, if it doesn't resolve it will wait forever.

Danny · Accepted Answer · 2022-09-19 08:58:48Z

0

You have a Promise that never resolves since you never call the resolve function. Here would be the fixed function.

function delay(ms = 0) {
  return new Promise((resolve, reject) => setTimeout(resolve ,ms))
}

answered Sep 19, 2022 at 8:58

Danny

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