如何专门化Javascript类？

Question

我在玩JavaScript类继承（我使用node.js）。我得到了子类的实例的“未定义”值。这里我的例子：

我定义了一个Squirrel类，我想在KillerSquirrel子类中专门化这个类。我想创建Squirrel和KillerSquirrel类的实例。

function Squirrel(name, color) { 
     this.name = name; 
     this.color = color; 
     console.log("A new " + this.color + " squirrel named " + this.name + " is born!!!"); 
    }; 

    // add a new method called speak() 
    Squirrel.prototype.speak = function(text) { 
     console.log(this.name + " squirrel says: '" + text + "'"); 
    }; 

    // specialize the Squirrel class by creating a new KillerSquirrel constructor 
    function KillerSquirrel(name, color) {  
     this.soul = 'very bad'; 
     console.log("Bhrrr ... a new " + this.color + " killer squirrel named " + this.name + " is born!!!"); 
    } 
    KillerSquirrel.prototype.__proto__ = Squirrel.prototype; 

    // add kill method 
    KillerSquirrel.prototype.kill = function(obj){ 
     console.log(this.name + " squirrel killed " + obj.name + " squirrel"); 
    } 

    // replace the speak() method 
    KillerSquirrel.prototype.speak = function(text) { 
         console.log(this.name + " squirrel says: 'Grhhh! " + text + "' Grhhh!"); 
    }; 

    var squirrel = new Squirrel("Gummy", "brown"); 
    squirrel.speak("My name is " + squirrel.name); 
    var killer = new KillerSquirrel("Mr.Hide", "black"); 
    killer.speak("My name is " + killer.name); 

我使用它的构造函数创建松鼠的松鼠实例，并将一些值传递给构造函数，并按预期工作。当我尝试使用它的构造函数创建一个KillerSquirrel类的实例并传递一些值时，杀手松鼠实例具有“未定义的属性”。

看到：（申请或电话）

$ node killersquirrel.js 
A new brown squirrel named Gummy is born!!! 
Gummy squirrel says: 'My name is Gummy' 
Bhrrr ... a new undefined killer squirrel named undefined is born!!! 
undefined squirrel says: 'Grhhh! My name is undefined' Grhhh! 

Answer 1

子类的构造应该调用父类的构造特殊结构手动，像这样：

function KillerSquirrel(name, color) { 
    Squirrel.apply(this, arguments);  
    this.soul = 'very bad'; 
    console.log("Bhrrr ... a new " + this.color + " killer squirrel named " + this.name + " is born!!!"); 
} 

或

function KillerSquirrel(name, color) { 
    Squirrel.call(this, name, color);  
    this.soul = 'very bad'; 
    console.log("Bhrrr ... a new " + this.color + " killer squirrel named " + this.name + " is born!!!"); 
} 

虽然此情况下（当参数相同时）第一种形式是优选的。