Instance returning undefined (JS)

Question

I am currently learning JavaScript from MDN and have run into a weird issue that I can't quite figure out. When I type newSquare.sideLength, it returns undefined, despite the instance having a number passed through it. What I need is for it to return the number 4 like I have passed through when creating the instance.

I've tried the same thing with a previous class and it worked perfectly fine but I don't quite understand what is wrong here.

Here's my code:

// OOJS 1
class Shape {
    name;
    sides;
    sideLength;
    constructor(name, sides, sideLength) {
        this.name = name;
        this.sides = sides;
        this.sideLength = sideLength;
    }

    calcPerimeter() {
        console.log(this.sides * this.sideLength);
    }
}

class Square extends Shape {
    constructor(sideLength) {
        super(sideLength);
        this.name = 'Square';
        this.sides = 4;
    }

    calcArea() {
        console.log(Math.pow(this.sideLength, 2));
    }
}

const newSquare = new Square(4);
console.log(newSquare.sideLength);

p.s. Sorry if this has a really simple solution, I've hit a block and I'm really trying to figure it out.

you aren't passing all 3 properties into Shape's constructor — Daniel A. White
– Daniel A. White, Commented Aug 18, 2022 at 13:40

Passionate Coder · Accepted Answer · 2022-08-18 13:43:56Z

3

Construction is just like another function and sequence of arguments should be same. when you pass super(sideLength); It sets name = 4 and other two arguments sets to undefined

You have to pass correct sequence of parameters in super function

super('Square', 4, sideLength);

answered Aug 18, 2022 at 13:43

Passionate Coder

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3 Comments

eraafay Over a year ago

Ah okay, thank you so much. Would've been helpful if MDN included solutions to their exercises but oh well. Thank you so much!

eraafay Over a year ago

Yes I will! I was trying to but it tells me I have to wait 5 minutes O.o

disinfor Over a year ago

I think the MDN for super is pretty explicit in this case: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… - super([arguments]) // calls the parent constructor. Just some helpful information for OP.

Carlos Osvaldo Chao Cortes · Accepted Answer · 2022-08-18 13:46:30Z

1

The Shape Class constructor, receive 3 properties, but in the square class, you only call the super constructor with 1 parameter, in this case, the name. So when you call newSquare.sideLength this property has not been initialized and its value is undefined. In Javascript, you need to respect the order of parameters in a constructor or function.

answered Aug 18, 2022 at 13:46

Carlos Osvaldo Chao Cortes

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