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JavaScript Object-Oriented JavaScript (2015) Constructor Functions and Prototypes Challenge Solution

nicholas maddren
nicholas maddren
12,793 Points
Posted by nicholas maddren
nicholas maddren
12,793 Points

Why did Andrew use this.sides?

In the dice function Andrew uses this.sides like so:

function Dice(sides) {
  this.sides = sides;
  this.roll = function() {
    var randomNumber = Math.floor(Math.random() * this.sides) + 1;
    return randomNumber;
  }
}

Why didn't he shorten his code by just including the sides parameter in the roll function? Like so:

function Dice(sides) {
  this.roll = function() {
    var randomNumber = Math.floor(Math.random() * sides) + 1;
    return randomNumber;
  }
}

I'm guessing there is a good reason behind this.

Thanks

3 Answers

Lars Reimann
Lars Reimann
11,816 Points

What you do here is called a closure. It means that even after the constructor function Dice has returned, the function this.roll still has access to the variables in the scope surrounding it when it was created, here sides. In this case I prefer your version. In general, however, you want to store any data directly in an instance and behavior in its prototype, which is shared across all instances of a constructor. Then you have to use Andrew's way:

function Dice(sides) {
    this.sides = sides;
}

Dice.prototype.roll = function() {
    var randomNumber = Math.floor(Math.random() * this.sides) + 1;
    return randomNumber;
}
Devjyoti Ghosh
Devjyoti Ghosh
2,422 Points
Devjyoti Ghosh
Devjyoti Ghosh
2,422 Points

Just to understand it properly, if this..sides is not set then during this example

var dice6 =  new dice(6);
var dice10 = new dice(10);

In dice10 the sides is still set at 6 and not changed to 10?

ammarkhan
seal-mask
.a{fill-rule:evenodd;}techdegree
ammarkhan
Front End Web Development Techdegree Student 21,661 Points
ammarkhan
.a{fill-rule:evenodd;}techdegree
ammarkhan
Front End Web Development Techdegree Student 21,661 Points

I am still learning but i think you cannot use sides on its own, you have to tell function that this.side to be in function is referring to sides passed in argument,or if you want to think of a car, all cars all gear but of you want to tell your car gear you will say javascript myCar.gear not toyota.gear, so you have to have reference to property or function i could be wrong.

nicholas maddren
nicholas maddren
12,793 Points
nicholas maddren
nicholas maddren
12,793 Points

It works by just passing the argument 'sides' straight into the function. Thanks for the input though.

Mike Ngu
Mike Ngu
8,229 Points
Mike Ngu
Mike Ngu
8,229 Points

It does work both ways, but it's a better habit doing it Andrew's way. There's this weird thing in JS where if you wrap another function inside of that function the this keyword will point to the global scope.