delaying the callback function in for loop without async/await

Asked Mar 23 '22 at 00:32

Active Mar 23 '22 at 00:54

Viewed 13 times

I have a function called printLetters which prints letters after 1000ms. I expected it to print letters one at a time with a 1000ms delay, but it prints all letters at once. How do I make it resolve each with delay without async/await?

const printLetters = (letters) => {
  for (let i = 0; i < logs.length; i++) {
    delay(1000).then(() => {
      console.log(logs[i]);
      })
  }
}

const delay = (t) => {
  return new Promise(res => setTimeout(res, t))
}

const arr = ['a','b','c']

printLetters(arr)

edited Mar 23 '22 at 00:54

asked Mar 23 '22 at 00:32

D Park

`delay(1000).then` means that all iterations will wait for 1 second, then all will resolve at the same time and log – CertainPerformance Mar 23 '22 at 00:33
@CertainPerformance thanks for your reply. Is there any way to make it resolve one at a time with a 1-second delay for each interval or is it just not possible? – D Park Mar 23 '22 at 00:37
See the canonical – CertainPerformance Mar 23 '22 at 00:38
@CertainPerformance Sorry but don't really see the answer I am looking for. So the delay with interval is only possible with async/await and not with .then()? – D Park Mar 23 '22 at 00:46
It's possible with `.then` (just multiply the timeout for each iteration), but `await` is a much, much cleaner approach – CertainPerformance Mar 23 '22 at 00:47
Could you show me the example code with `.then()`? – D Park Mar 23 '22 at 00:50

delaying the callback function in for loop without async/await

0 Answers0