setTimeout Behaviour with two different statements

Question

    setTimeout(()=> alert('hi'), 3000); 
    setTimeout(()=> alert('hello'), 5000);

I am really struggling hard understanding setTimeout method. In the above code, the first statement works correctly meaning, it displays hi after 3 seconds.

But the next statement doesn't delay by 5seconds. Why? The second output is shown immediately after the first one without any dealy. Please explain the behaviour. Refer me some article to understand in detail.

Neither of your `setTimeout()` calls will delay at all, because you're **calling** `alert()` before `setTimeout()`. — Pointy, Jun 29 '20 at 13:38
and the point that the method call is missing an `e` (time*_e_*out not timout) — Jamiec, Jun 29 '20 at 13:39
To use function with parameter inside `setTimeout` use like this way `setTimeout(alert, 3000, 'hi'); `. — Karan, Jun 29 '20 at 13:46
@thelostcause What you want is `setTimeout (() => (alert ('hi'), setTimeout (() => alert ('hello'))))` — Matías Fidemraizer, Jun 29 '20 at 13:47
"*The second output is shown immediately after the first one without any delay*" - only when you take more than 2s to confirm the first alert. — Bergi, Jun 29 '20 at 14:35

score 0 · Answer 1 · answered Jun 29 '20 at 14:48

Actually it is working correct:

console.log('start now:',new Date().toISOString());
setTimeout(()=> console.log('hi now:',new Date().toISOString()), 3000); // 3 seconds after starts
setTimeout(()=> console.log('hello now:', new Date().toISOString()), 5000); // 5 seconds after starts, 2 seconds after hi.

setTimeout Behaviour with two different statements

1 Answers1