这种Javascript继承方法有什么问题吗

好吧，这里有一件事是错误的：

var dog = new Dog();
console.log(dog instanceof Dog) //=> false

问题是，您所做的实际上是而不是继承。您只是简单地使用函数来返回动物对象的修改版本。这导致了一种类似于继承但又不同的行为。

JavaScript的对象模型使用称为原型链的东西，这与典型的面向对象继承非常相似。Mozilla开发者网络在这方面有一篇很棒的文章：继承和原型链。

以下是如何在示例中使用原型继承的示例：

var Animal = function() {};
Animal.prototype = {
    legs: 4,
    makeNoise: function () {
        return "noise";
    }
};
var Dog = function () {};
// Notice: Dog.prototype is inheriting from Animal.prototype
Dog.prototype = Object.create(Animal.prototype);
Dog.prototype.makeNoise = function () {
    var noise = "bark 'n";
    // Notice: calling Animal's makeNoise function with
    // 'this' set to our dog object.
    noise += Animal.prototype.makeNoise.call(this);
    return noise;
};
var Monkey = function () {};
// Notice: Monkey.prototype is inheriting from Animal.prototype
Monkey.prototype = Object.create(Animal.prototype)
Monkey.prototype.legs = 2;
Monkey.prototype.makeNoise = function () {
    var noise = "weird monkey sound 'n";
    // Notice: calling Animal's makeNoise function with
    // 'this' set to our monkey object.
    noise += Animal.prototype.makeNoise.call(this);
    return noise;
};
$("#monkey").click(function () {
    var monkey = new Monkey();
    var result = "Legs: " + monkey.legs + "'n";
    result += "Noise: " + monkey.makeNoise();
    alert(result);
});
$("#dog").click(function () {
    var dog = new Dog();
    var result = "Legs: " + dog.legs + "'n";
    result += "Noise: " + dog.makeNoise();
    alert(result);
});
$("#inheritance").click(function () {
    var dog = new Dog();
    var monkey = new Monkey();
    var result = 'dog instanceof Dog = ' + (dog instanceof Dog) + "'n" +
        'dog instanceof Animal = ' + (dog instanceof Animal) + "'n" +
        'dog instanceof Monkey = ' + (dog instanceof Monkey) + "'n" +
        'monkey instanceof Monkey = ' + (monkey instanceof Monkey) + "'n" +
        'monkey instanceof Animal = ' + (monkey instanceof Animal) + "'n" +
        'monkey instanceof Dog = ' + (monkey instanceof Dog) + "'n";
    alert(result);
});

查看JS Fiddle

这是一篇优秀的文章：http://javascriptissexy.com/oop-in-javascript-what-you-need-to-know/

//Constructor/Object
function Animal(inNoise) {
    this.legs = 4;
    this.noise = inNoise;
}
//Set Objects prototypes
Animal.prototype = {
    constructor: Animal,
    changeNoise: function(newNoise){this.noise = newNoise;}
}
//Create some dogs
var dog = new Animal('bark');
var pitbull = Object.create(dog);
var chiahua = Object.create(dog);
//Change dogs noise
chiahua.changeNoise('littleBark');
alert(dog.noise); //bark
alert(pitbull.noise);//bark
alert(chiahua.noise); //littlebark
alert(chiahua.legs); //4

除了返回self会破坏继承链之外，您可能对JavaScript的继承技术有误解：

JavaScript是一种基于原型的语言，这意味着正确的继承是这样设置的：

function Animal(){}
Animal.prototype.walk = function(){
    console.log('walking...');
};
Animal.prototype.roar = function(){
    alert('roar');
};
function Dog(){}
Dog.prototype = new Animal();
Dog.prototype.roar = function(){
    alert('bark!');
};
var animal = new Animal();
var dog = new Dog();
animal.walk();
dog.walk();   // both can call the walk function
animal.roar(); // alerts roar
dog.roar(); // alerts bark

原型继承允许共享对象定义的内存(将一个对象记录到控制台以了解我的意思(。换句话说，Dog的每个实例都将使用相同的roar函数，但通过在构造函数中将属性直接附加到this，可以防止原型链，并将函数的内存委派给该对象的每个实例。如果你要创建该对象的许多实例，应用程序的性能将反映这一点。

@在我的例子中，Thom正在展示经典继承与原型。两者都是对的，但这里有一个很好的Stack答案。