setTimeout inside a function [duplicate]

Question

function delay2sec(){
  let name = "abc"
  setTimeout(function () {
    name = "xyz"
  }, 0)

  return name
}

console.log(delay2sec())

The output is abc.

Can someone explain why it is not reassigning the name to 'xyz'?

Can someone please explain why it is not reassigning the name to 'xyz'? It does reassign it, but that code happens after everything else, so it's not very useful. You create name = 'abc', then you set a timeout, then you return 'abc', then you log 'abc'. Later on, the timeout goes off and you assign 'xyz'. — Nicholas Tower
– Nicholas Tower, Commented Aug 28, 2022 at 14:45
I think the problem here is likely not understanding that Javascript is mostly synchronous, and non-blocking, so the code execution doesn't wait for that setTimeout to resolve, and instead continues to the return - This is what async / await is used for -- There should be lots of resources to explain this concept, but here's the first I found to point in the right direction freecodecamp.org/news/synchronous-vs-asynchronous-in-javascript — Shiny
– Shiny, Commented Aug 28, 2022 at 14:48

Lajos Arpad · Accepted Answer · 2022-08-28 14:51:48Z

0

In order to understand this behavior, you need to understand how the event loop works:

Your setTimeout adds an event to the event loop, which will be executed after the current function. Your return statement is evaluated before the current function ends, so the return evaluates the value before the change.

answered Aug 28, 2022 at 14:51

Lajos Arpad

80.4k42 gold badges123 silver badges235 bronze badges

Sign up to request clarification or add additional context in comments.

Collectives™ on Stack Overflow

setTimeout inside a function [duplicate]

1 Answer 1

Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Linked

Related