How so I stop a shared method from running with class and extends?

July 5, 2022

I am creating two classes, Rectangle and Square, based on the example shown on MDN.

Since Rectangle is the prototype of Square, they share the same method area().

However, I want area() to run only if length === height for the square.

I’ve tried to add an if-statement after super(), but it area() still returns this.length* this.height regardless of an erroneous input.

I could add an if-statement in the area() method, but I did not because it would not make sense when we are calculating the area of a rectangle.

class Rectangle {
  constructor(length, height){
    this.length = length;
    this.height = height;
  }
  area(){
    return this.length*this.height;
  }
}

class Square extends Rectangle {
  constructor(length, height){ // have area() return "Not a square" instead of a number
    super(length, height);
    if (length !== height) {
      console.log("Not a square.")
    }
    this.name = "square";
  }
}

const rectangle = new Rectangle(2,3);
console.log(rectangle.area());

const square = new Square(5,2);
console.log(square.area())

>Solution :

an idea can be to overload area method in rectlangle to only call rectangle method when length === height

 area() {
    return (this.length === this.height) ? super.area() : undefined;
}

class Rectangle { 
 constructor(length, height){
    this.length = length;
    this.height = height;
  }
  area(){
    return this.length*this.height;
  }
}

class Square extends Rectangle {
  constructor(length, height){ // have area() return "Not a square" instead of a number
    super(length, height);
    if (length !== height) {
      console.log("Not a square.")
    }
    this.name = "square";
  }
  
  area(){
    return (this.length === this.height) ? super.area() : undefined;
  }
}

const rectangle = new Rectangle(2,3);
console.log(rectangle.area());

const square = new Square(5,2);
console.log(square.area())