为什么我的子类没有从 Javascript 中的父类继承属性？

Question

我正在尝试了解 Javascript 的原型继承模型，但似乎缺少某些东西。 我创建了以下对象集：

 function Dog() { this.legs = 4; this.arms = 0; } Dog.prototype.describe = function() { console.log("Has " + this.legs + " legs and " + this.arms + " arms"); } function Schnauzer(name) { Dog.call(this); this.name = name; } Schnauzer.prototype.describe = function() { console.log(this.name + " is a Schnauzer with a shaggy beard"); Dog.prototype.describe(this); } var my_dog = new Schnauzer("Rupert"); my_dog.describe();

我的期望是这将输出：

Rupert is a Schnauzer with a shaggy beard
Has 4 legs and 0 arms

然而，这是我实际得到的：

Rupert is a Schnauzer with a shaggy beard
Has undefined legs and undefined arms

Answer 1

和

Dog.prototype.describe(this);

您正在使用Dog.prototype的调用上下文（ this值）调用describe方法，并将 dog 实例作为第一个参数传递。 但是describe不接受任何参数，它试图从this获取属性。

做Dog.prototype.describe.call(this); 相反，要使用 dog 实例的this值调用该方法：

 function Dog() { this.legs = 4; this.arms = 0; } Dog.prototype.describe = function() { console.log("Has " + this.legs + " legs and " + this.arms + " arms"); } function Schnauzer(name) { Dog.call(this); this.name = name; } Schnauzer.prototype.describe = function() { console.log(this.name + " is a Schnauzer with a shaggy beard"); Dog.prototype.describe.call(this); } var my_dog = new Schnauzer("Rupert"); my_dog.describe();

尽管如此，这还是有点难看，因为您必须使用.call来获取一个方法，在正常情况下，该方法在当前对象的原型链上会更高。 如果可能，您可以考虑重新命名您的Schnauzer.prototype.describe方法，这样它就不会掩盖Dog.prototype方法：

 function Dog() { this.legs = 4; this.arms = 0; } Dog.prototype.describe = function() { console.log("Has " + this.legs + " legs and " + this.arms + " arms"); } function Schnauzer(name) { Dog.call(this); this.name = name; } Schnauzer.prototype = Object.create(Dog.prototype); Schnauzer.prototype.describeSchnauzer = function() { console.log(this.name + " is a Schnauzer with a shaggy beard"); this.describe(); } var my_dog = new Schnauzer("Rupert"); my_dog.describeSchnauzer();

或者使用class语法和super ：

 class Dog { legs = 4; arms = 0; describe() { console.log("Has " + this.legs + " legs and " + this.arms + " arms"); } } class Schnauzer extends Dog { constructor(name) { super(); this.name = name; } describe() { console.log(this.name + " is a Schnauzer with a shaggy beard"); super.describe(); } } var my_dog = new Schnauzer("Rupert"); my_dog.describe();