Q.denodeify example not executing properly

Question

I am a newbie in node just trying out an example on the usage of Q. I am not able to get the log 'addition completed' printed. I am getting 'inside 7' & '7' in the output. What am I doing wrong?

My code is

var Q = require('q');

function Add(callback){

    setTimeout(function(){
        var a, b, c;
        b = 3; 
        a = 4;
        c = a + b; 
        console.log("inside "+c);
        callback(c);
    }, 2000); 

}

var promise = Q.denodeify(Add);

function callback(x){
    console.log(x);
}

promise(callback).then(function(){
    console.log('addition completed');
});

ok. i just posted code that works but i need an explanation for it. — jet nebula, Feb 10 '20 at 07:34
Thank you for your answer. i was trying out a tutorial example online given for Q using code similar to one i posted and wanted to learn how to do so. i understand q is not required for this example. — jet nebula, Feb 10 '20 at 07:41

Cat · Answer 1 · 2020-02-10T07:28:36.997

This sample code won't answer any questions you might have about the q library.
Instead, it just demonstrates how you can use a promise instead of a callback to complete an asynchronous task -- which seems like the underlying goal of your code.

(For more info and examples, you can check out MDN: Promises)

Note that promises are an alternative to callbacks, so an expression like promise(callback) is not something you'd typically write.

Also, I only looked at q very briefly, but I think Q.denodeify might not be the correct method to use for the task you're trying to accomplish.

For suggestions specific to q, you might also find this answer helpful.

function add(){

  // Calling `add` will return a promise that eventually resolves to `7`
  // There is no way this particular promise can reject (unlike most promises)
  const myPromise = new Promise(function(resolve, reject) {

    // The anonymous function we pass to `setTimeout` runs asynchronously
    // Whevever it is done, it applies the promise's `resolve` function to `c`
    setTimeout(function(){
      const
        a = 4,
        b = 3,
        c = a + b;
      resolve(c);
    }, 200);
  });
  return myPromise;
}

// `add` returns a promise that always resolves to `7`
// The `.then` method of this promise lets us specify a function that 
//   does something with the resolved value when it becomes available
add().then(function(resolvedValue){
  console.log("addition result: " + resolvedValue);
});

score 0 · Answer 2 · answered Feb 10 '20 at 07:33

I tweaked it a bit and got it to work. But i am not able to understand clearly why this worked and the one prior to this failed. Hope someone call help with an explanation.

My new code is

var Q = require('q');

function Add(callback){

    setTimeout(function(){
        var a, b, c;
        b = 3; 
        a = 4;
        c = a + b; 
        console.log("inside "+ c);
        callback();

    }, 2000); 

}

var promise = Q.denodeify(Add);

promise().then(function(){
    console.log('addition completed');
});

score 0 · Answer 3 · answered Feb 10 '20 at 07:38

you can use this example for the promise

 var Q = require('q');

    function add(callback){

        return new Promise(function ( resolve, reject) {
            setTimeout(function(){
                var a, b, c;
                b = 3;
                a = 4;
                c = a + b;
                console.log("inside "+c);
                callback(c);
                if(c){
                    resolve(c);
                }else{
                    reject();
                }
            }, 2000);

        });
    }

    //var promise = Q.nbind(add);

    function callback(x){
        console.log(x);
    }

    add(callback).then(function(){
        console.log('addition completed');
    }).catch(function (E) {
        console.log(E);
    });

Q.denodeify example not executing properly

3 Answers3