Correct way to implement a Factory pattern in JavaScript in the following code

Question

The following code is supposed to abstract a temperature sensor for a project involving Arduino.

I have the Sensor function representing the hardware sensor, a sensorFactory function to create instances of the sensor, and a saveSensor / findSensor function that stores / returns the sensor objects from an array.

var sensors = [];
const Sensor = (sensorId, sensorType, sensorName) =>{
    var temperature;
    let sensor = {
        sensorId,
        sensorType,
        sensorName,
        temperature,
        readTemperature: (temp) => {
            temperature = temp //simulates reading from hardware
        },
        getTemperature: () => {
            console.log(temperature)
        }
    }
    return sensor; 
};
const SensorFactory = () =>({
    createSensor: (sensorId, sensorType, sensorName) => {
        var sensor = {};
        switch(sensorType){
            case "DS18B20":         
                sensor = Sensor(sensorId,sensorType, sensorName);
                saveSensor(sensor);
                return findSensor(sensorId);
                break;
        }
    },
    getSensor: (sensorId) =>{
    if (sensorId){
            return findSensor(sensorId);            
        }
        return sensors;
    }
});
const saveSensor = (sensor) => {
    sensors.push(sensor);
};
const findSensor = (sensorId) => {
    return sensors[sensorId -1];
};
module.exports = SensorFactory;

I have two questions:

Why can't I see the value of the temperature variable when I run the line console.log(JSON.stringify(sensor1))?

const SensorFactory = require ("./sensor.js");
var sensorFactory   = SensorFactory();
sensorFactory.createSensor (1,"DS18B20", "Cryogenic Tank");

var sensor1 = sensorFactory.getSensor(1);
sensor1.readTemperature(200);
sensor1.getTemperature(); //prints 200  
sensor1.readTemperature(100); //prints 100
sensor1.getTemperature();   
sensor1.readTemperature(10); //prints 10
sensor1.getTemperature();

console.log(JSON.stringify(sensor1))

Expected:

{"sensorId":1,"sensorType":"DS18B20","sensorName":"Cryogenic Tank", "temperature": 10}

Actual output:

{"sensorId":1,"sensorType":"DS18B20","sensorName":"Cryogenic Tank"}

I believe that temperature (A) in the following block is a private variable so it should not be accessible from outside the object, but shouldn't I be able to see the temperature variable (B) when I run console.log(JSON.stringify(sensor1))?

const Sensor = (sensorId, sensorType, sensorName) =>{
    var temperature; <----A----private. 
    let sensor = {
        sensorId,
        sensorType,
        sensorName,
        temperature, <----B----I should be able to see this, right?
        readTemperature: (temp) => {
            temperature = temp;
            console.log('readTemperature');             
        },
        getTemperature: () => {             
            console.log('getTemperature');
            console.log(temperature);           }
    }
    return sensor;
};

I am very rusty with JS. Can you please help me understand where I am conceptually wrong? How would you modify the code so that it works as expected?

Why is SensorFactory a function that returns a factory object? It would seem that all factories are the same. Just use module.exports = {createSensor(…) { … }, getSensor(…) { … }}; — Bergi
– Bergi, Commented Apr 20, 2021 at 19:34
"Why can't I see the value of the temperature variable" - because JSON.stringify only prints the value of the .temperature object property, not the temperature local variable that your readTemperature method assigns. Use getTemperature() instead and remove the immutable property from your object - or remove the variable from your code and always access the property. — Bergi
– Bergi, Commented Apr 20, 2021 at 19:36
@Bergi this was a big part the solution for me. I was trying to be lazy and see that readTemperature was changing the value of the temperature variable before passing the object to the handlebars template. Also I forgot to add the return temperature line in the getTemperature method. My code works now as intended. Can you help me understand what do you mean when you say "Why is SensorFactory a function that returns a factory object?" My idea is to have a SensorFactory returning different types of sensor objects. — acanessa
– acanessa, Commented Apr 22, 2021 at 6:16
"My idea is to have a SensorFactory returning different types of sensor objects." - yes, a SensorFactory should do that, like the concrete sensorFactory does with its createSensor method. But why do you have a SensorFactory function to create the factory? — Bergi
– Bergi, Commented Apr 22, 2021 at 8:12
@Bergi: Because I am not a programmer, so some concepts are confusing for me. I understand what you are asking now, in fact I am following your advice in another module and exporting as you have suggested. — acanessa
– acanessa, Commented Apr 26, 2021 at 2:29

georg · Accepted Answer · 2021-04-20 18:29:06Z

1

Here's some refactoring into modern Javascript (answers the second question, and, as a side effect, also the first ;)

class Sensor {
    constructor(id, type, name) {
        this.id = id
        this.type = type
        this.name = name
        this.temperature = null
    }

    readTemperature(temp) {
        this.temperature = temp //simulates reading from hardware
    }
}


class SensorSet extends Map {
    add(id, type, name) {
        this.set(id, new Sensor(id, type, name))
        return this.get(id)
    }
}

let sensors = new SensorSet()

sensors.add(1, "DS18B20", "Cryogenic Tank")

let s1 = sensors.get(1)

s1.readTemperature(200)
console.log(s1.temperature)

s1.readTemperature(100)
console.log(s1.temperature)

s1.readTemperature(10)
console.log(s1.temperature)

console.log(JSON.stringify(s1))

answered Apr 20, 2021 at 18:29

georg

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4 Comments

acanessa Over a year ago

Thank you. I should have clarified that I was trying to avoid using classes. Apologies. How would you suggest to do this only using functions?

georg Over a year ago

"avoid using classes." - why?

acanessa Over a year ago

I had pass binding issues using classes, and I've read that I should favour functions instead of classes i.e. link, and function composition instead or inheritance i.e. link. Your code is clear and is how I used to code in PHP, it makes a lot of sense. However, I've found using functions and composition easier in the past than using classes and inheritance. I will continue working on this and post my updates.

georg Over a year ago

@acanessa: I see. Unfortunately, internet is full of misleading if not outright wrong stuff (re the first link). My advice would be stick to whatever feels more comfortable to you. Good luck with your project!

acanessa · Accepted Answer · 2021-04-26 02:34:54Z

0

@Bergi's second comment is the solution to my question. I have also followed his advice and eliminated the sensorFactory function that returns the sensorFactory object. Code is much clear now.

answered Apr 26, 2021 at 2:34

acanessa

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