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Is there any differences between "this" and "super" when accessing parent class' property?

I always think "super" is the only way to access parent class' property, but i realized that "this" can also access parent class' property. Is there any differences betweet "super" and "this" in this situation?

class Human {
  constructor (name, age) {
    this.name = name;
    this.age = age;
  }
  mouse (sentense) {
    console.log(`${this.name}: "${sentense}".`);
  }
}

class Footballer extends Human {
  constructor(name, age, position, team) {
    super(name, age);
    this.team = team;
    this.position = position;
  }
  talk (sentense) {
    super.mouse(sentense); // This works.
    this.mouse(sentense); // This also works.
  }
}
about 4 years ago · Santiago Gelvez
2 Respuestas
Responde la pregunta

0

If you use this, the engine will first try to access the property on the instance itself. If it doesn't exist on the instance, it'll then go to the prototype (Footballer.prototype). If it doesn't exist there, it'll then go to Human.prototype to see if the property exists.

If you use super, it'll immediately go to the parent class, Human.prototype, to see if the property exists, and will go upward from there (rather than go through the instance and child prototype first).

This will be especially important to keep in mind if a method with the same name exists both on the parent and child, which isn't uncommon.

class Human {
  mouse () {
    console.log('parent class');
  }
}

class Footballer extends Human {
  mouse() {
    console.log('child class');
  }
  talk () {
    super.mouse();
    this.mouse();
  }
}

const f = new Footballer();
f.talk();

about 4 years ago · Santiago Gelvez Denunciar

0

There is a big difference. In languages like Java and Javascript, with super you are accessing the parent's properties and methods.

With this you are accessing the child's properties and methods.

this.mouse() works because first the mouse property is accessed for the child (Footballer) and when not found the property is looked for in the parent (Human). This is how inheritance works. When accessing a property called x, Javascript will try to move upwards in the prototype chain until it finds a property named x.

If you define a property like kick only in Footballer, then super.kick() will not work but this.kick() will, demonstrating the difference.

class Human {
  constructor (name, age) {
    this.name = name;
    this.age = age;
  }
  mouse (sentense) {
    console.log(`${this.name}: "${sentense}".`);
  }
}

class Footballer extends Human {
  constructor(name, age, position, team) {
    super(name, age);
    this.team = team;
    this.position = position;
  }
  kick(){
  console.log("kicked");
  }
  play(){
    this.kick();
    super.kick();
  }
  talk (sentense) {
    super.mouse(sentense); // This works.
    this.mouse(sentense); // This also works.
  }
}

const messi = new Footballer();
messi.play();

about 4 years ago · Santiago Gelvez Denunciar
Responde la pregunta
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