Get return value from setTimeout [duplicate]

javascript return settimeout return-value

This question already has answers here: How do I return the response from an asynchronous call? (43 answers) Closed 2 years ago.

I just want to get the return value from setTimeout but what I get is a whole text format of the function?

function x () {
    setTimeout(y = function () {
        return 'done';
    }, 1000);
    return y;
}

console.log(x());

your syntax is like that, it will return function only.

do you want "done" to be returned? How can you return a result of function, which will be called 1000 ms later?

Timeouts are asynchronous, so you can't return from them. (well, you can, but the return value is going nowhere)

better to get a callback in function x and call that function with done.

const x = async (waitMs) => new Promise((resolve) => setTimeout(() => resolve('done'), waitMs)); console.log(await x(1000));

rajniszp

You need to use Promises for this. They are available in ES6 but can be polyfilled quite easily:

function x() {
   return new Promise((resolve, reject) => {
     setTimeout(() => {
       resolve('done!');
     });
   });
}

x().then((done) => {
  console.log(done); // --> 'done!'
});

With async/await in ES2017 it becomes nicer if inside an async function:

async function() {
  const result = await x();
  console.log(result); // --> 'done!';
}

actually the done in my example was a promise that needs to be returned inside a setTimeout..because i cant get the returned deferred.resolve() inside setTimeout. how would it be? thank you

Can you post a clearer example of your code in the question?

Quentin

You can't get a return value from the function you pass to setTimeout.

The function that called setTimeout (x in your example) will finish executing and return before the function you pass to setTimeout is even called.

Whatever you want to do with the value you get, you need to do it from the function you pass to setTimeout.

In your example, that would be written as:

function x () {
    setTimeout(function () {
        console.log("done");
    }, 1000);
}

x();

Mritunjay

Better to take a callback for function x and whatever task you want to perform after that timeout send in that callback.

function x (callback) {
    setTimeout(function () {
        callback("done");
    }, 1000);
}

x(console.log.bind(console)); //this is special case of console.log
x(alert)

Bare in mind, that this will not await after the function call, so it might actually not happen. For example add process.exit, after calling x and "done" will be never done.

Satyabrata Saha

You can use a combination of Promise and setTimeOut like the example below

let f1 = function(){
    return new Promise(async function(res,err){
        let x=0;
        let p = new Promise(function(res,err){
            setTimeout(function(){
                x= 1;
                res(x);
            },2000)
        })
        p.then(function(x){
            console.log(x);
            res(x);
        })


    });
}

tuan huynh

I think you want have flag to know event occured or no. setTimeout not return a value. you can use a variable to detect event occured or no

var y="notdone";
   setTimeout(function () {
         y="done";
    }, 1000);

You can access variable y after timeout occured to know done or not

Get return value from setTimeout [duplicate]

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