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JavaScript

Posted by Erwan EL
Erwan EL
7,669 Points

Why using .this into the roll method

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


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

Tried both with various dices and both works.

2 Answers

Steven Parker
Steven Parker
243,318 Points
Steven Parker
Steven Parker
243,318 Points

In the first, you establish an instance variable named "sides" and use it in the "roll" method.

In the second, you establish but then ignore the instance variable. The "roll" method is using the argument that was passed in directly.

It might not seem different until you try to reconfigure the object:

var a = new Dice1(6);  // this is the first version, just renamed to be unique
var b = new Dice2(6);  // the second version
a.roll();              // this gives a number from 1-6
b.roll();              // this also gives a number from 1-6 - the same, so far
a.sides = 100;         // now we reconfigure for 100 sides
b.sides = 100;         // same here (it seems)
a.roll();              // now this gives a number from 1-100
b.roll();              // but this STILL gives a number from 1-6 !

Since you don't use the "this.sides", you could actually condense the second function to just this:

function Dice(sides) {
    this.roll = () => Math.floor(Math.random() * sides) + 1;
}
Erwan EL
Erwan EL
7,669 Points
Erwan EL
Erwan EL
7,669 Points

Thanks a lot, it's very clear now.