Async function called with call or apply never resolved

Question

Functions that are called with call or apply using await never resolve, please check the following code snippet

 const obj = { resolveAfter2Seconds: function() { return new Promise(resolve => { setTimeout(() => { resolve('resolved'); }, 2000); }); } } async function asyncCall() { console.log('calling'); var result = await obj.resolveAfter2Seconds(); console.log(result);//resolved var result2 = await obj.call("resolveAfter2Seconds"); console.log(result2);//never alled } asyncCall();

Answer 1

Just expanding @CertainPermances comment - call is a function method used to assign an option to this inside the function, you can't use it on an object

(function() {
    const obj = {
        resolveAfter2Seconds : function() {
            return new Promise(resolve => {
                setTimeout(() => {
                resolve('resolved');
                }, 2000);
            });
        }
    }

    async function asyncCall() {
        console.log('calling');
        var result = await obj.resolveAfter2Seconds();
        console.log(result);//resolved

        // var result2 = await obj.call("resolveAfter2Seconds");    caused an eror
        var result2 = await obj.resolveAfter2Seconds();
        // or
        var result2 = await obj.resolveAfter2Seconds.call( obj /* perhaps */); // which makes no difference at all because this = obj anyway
        console.log(result2);
    }

    asyncCall();

})();

Answer 2

I was asking the question in wrong way, I ended up using the correct syntax as follows

const obj = {
   resolveAfter2Seconds: function() {
    return new Promise(resolve => {
      setTimeout(() => {
        resolve('resolved');
      }, 2000);
    });
  }
}

async function asyncCall() {
  console.log('calling');
  var result = await obj.resolveAfter2Seconds();
  console.log(result);//resolved

 var result2 = await obj["resolveAfter2Seconds"].call();
  console.log(result2);//never alled
}

asyncCall();