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Get result of currently running async function

How to check if there is already running function and if it is exist listen to this function result;

async function a() {
// wait 5 seconds and return foo = foo + 1;
// if A is already running await and return result of this running function A instead result;
}
about 4 years ago · Juan Pablo Isaza
3 Respuestas
Responde la pregunta

0

If I translate the problem correctly, A returns a promise that is asynchronously settled. While the promise is pending, all calls to a wrapper function around A should return the currently pending promise.

If, however, A has not been called, or a previously return promise has been settled, A should be called again.

This can be achieved by chaining off the promise returned by A, using promise handlers to determine results are no longer pending, and have the wrapper function return the chained promise. This example code speeds up the process a little - four successive calls made to a 500ms apart get the same fulfilled value from A which is taking 2000ms to perform a mythical asynchronous task:

// promise a delay

const delay = ms => new Promise(resolve=>setTimeout(resolve, ms));

// async function A

let foo =0;
async function A() {
  await delay( 2000); // 2 second example
  return foo = foo + 1;
}

// Wrapper function a

const a=(()=>{ 
  let pending = null;
  const onfulfill = data => { pending = null; return data};
  const onreject = err => { pending = null; throw err};
  let a = ()=>  pending || (pending = A().then(onfulfill, onreject));
  return a;
})();

// and test

async function test() {
  for( let i=1; i < 11; ++i) {
    a().then( data=> console.log(`a() call ${i} fulfills with ${data}`));
    await delay(500);
  }
}
console.log(" a is a named function ", a.name == 'a')
test();

  • a is coded to be a named function which minimizes run time object creation by using the two parameter form of then and passing pre-compiled functions as handler arguments.
  • The catch handler re-throws the error for caller code to handle.
  • synchronous and asynchronous code don't mix well together. Calls to wrapper function a always receive a pending promise value in return.
about 4 years ago · Juan Pablo Isaza Denunciar

0

Solution
You can create class that will be execute one flow (maximum) and await result if flow already running. It may looks something like that:

class OneThreadExecutor {

    // Boolean variable which represents is task running right now
    taskRunning = false;

    // All Promise.resolve callbacks
    listeners = [];

    // Accept initial value
    constructor(value = 0) {
        this.value = value;
    }

    // Send [result = value + 1] after 5 sec
    startNewTask = () => {
        this.taskRunning = true;
        setTimeout(() => {
            this.taskRunning = false;
            return this.sendResult();
        }, 5000)
    }

    // Call all Promise.resolve callbacks, and pass [value + 1] to them
    sendResult = () => {
        for (const listener of this.listeners) {
            listener(++this.value);
        }
        this.listeners = [];
    }

    // Main method that exec one task
    getResult = () => new Promise(resolve => {
        // Add callback to queue
        this.listeners.push(resolve); 
        // Start new task if necessary
        if (!this.taskRunning) this.startNewTask(); 
    })

}

General concept
Async getResult method will register promise in class' queue. Any successfull task execution will send result to queue. Current task returns value + 1 on each getResult call and take 5 seconds for whole flow.

Usage

const a = new OneThreadExecutor(); // value = 0 by default

// will start task#1, 5 sec left till .then call, value = 1
a.getResult().then(...)
// 2.5 sec left, task#1 is already half progress, value = 2 
setTimeout(() => {
    a.getResult().then(...) 
}, 2500)

Async/await


const a = new OneThreadExecutor(3); // let's set value = 3 initially

// will start task#1, 5 sec left till .then call, value = 4
a.getResult();
// wait till task#1 completed, 5 sec left, value = 5 
const value = await a.getResult();
// will start task#2 bacause not task running, 5 sec left, value = 6 
a.getResult();

Cons
In demo solution we already expect successful task execution, without error handling, so you may need to extend it for proper error catching during task execution.

about 4 years ago · Juan Pablo Isaza Denunciar

0

var is_a_running = false;
function a() {
    if (is_a_running == false) {
        is_a_running = true
        //do something
        //after you have done the thing
        is_a_running = false
    }
    else if (is_a_running == true){
        result();
    }
}

This is should help you

about 4 years ago · Juan Pablo Isaza Denunciar
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