javascript - 为什么我无法在JS继承中从子类实例访问父类的属性？

Question

我有两个类。我想从实例访问 Parent 的 type 属性:

// Parent class
function Animal() { this.type = 'animal' }

// Child class
function Rabbit(name) { this.name = name }

// I inherit from Animal
Rabbit.prototype = Object.create(Animal.prototype);
Rabbit.prototype.constructor = Rabbit; // I want to keep Rabbit constructor too

// I instantiate my Rabbit and am trying to access rabbit.type
const rabbit = new Rabbit('Bunny');
rabbit.name // => Bunny
rabbit.type // => undefined. WHY?

我知道如何解决它并访问类型，但是...

// all is the same

// Child class
function Rabbit(name) {
  Animal.apply(this, arguments); // Just need to add this line in Rabbit class
  this.name = name 
}

// all is the same

rabbit.name // => Bunny
rabbit.type // => animal

...但是为什么它在第一个示例中不起作用？不使用Animal.apply是否可以实现它？

Answer 1

是的，如果您将type添加到原型(prototype)中:

  Animal.prototype.type = "animal";

或者您可以将 Animal.apply 调用隐藏在 class 糖后面:

 class Animal {
   constructor() {
      this.type = "animal";
   }
 }

 class Rabbit {
   constructor(name) {
     super(); // <<<
     this.name = name;
   }
 }