JavaScript实现继承
以下内容均基于本人对《JavaScript高级程序设计》第三版6.3小节的理解
先看一下父类
function Animal(name) { var name = name; //'私有（受保护）'成员，只允许在父类的构造函数中赋值 this.food = undefined; //'公有'成员 //引用类型的成员 this.birthday = { year: undefined }; //构造函数中的方法，打印一些基本信息 this.greeting = function() { console.log('Hi, my name is {' + name + '} I like eat {' + this.food + '} and my birth year is {' + this.birthday.year + '}'); }; //原型中的方法，将构造函数中的非函数成员以JSON格式打印 Animal.prototype.briefInfo = function() { var brief = { name: name, food: this.food, birthday: this.birthday }; console.log(JSON.stringify(brief)); }; }方式一：原型链继承
实现方式：子类的原型指向父类的实例
子类：
function Dog() {}Dog.prototype = new Animal(); //原型指向父类的实例测试：
var dog1 = new Dog();dog1.food = 'shit';dog1.birthday.year = 2015;var dog2 = new Dog();dog2.food = 'bones';dog2.birthday.year = 2016;dog1.greeting(); //console: Hi, my name is {undefined} I like eat {shit} and my birth year is {2016}dog2.greeting(); //console: Hi, my name is {undefined} I like eat {bones} and my birth year is {2016}dog1.briefInfo(); //console: {"food":"shit","birthday":{"year":2016}}dog2.briefInfo(); //console: {"food":"bones","birthday":{"year":2016}}//以上，//birthday是引用类型的属性，所以dog1的birthday.year被dog2覆盖了;//无法给dog1和dog2的name赋值存在的问题：
引用类型的对象会被子类的所有实例共享(1.1)
无法在创建子类的实例时，给父类的构造函数传递参数(1.2)
方式二：借用构造函数（伪造对象、经典继承）
实现方式：在子类的构造函数中利用call(或者apply)方法执行父类构造函数(问题1.2解决)，将执行对象设为子类的this,相当于把父类构造函数中的成员拷贝了一份到子类(问题1.1解决)
子类
function Dog(name) { Animal.call(this, name);}测试
var dog1 = new Dog('tom');dog1.food = 'shit';dog1.birthday.year = 2015;var dog2 = new Dog('mike');dog2.food = 'bones';dog2.birthday.year = 2016;dog1.greeting(); //console: Hi, my name is {tom} I like eat {shit} and my birth year is {2015}dog2.greeting(); //console: Hi, my name is {mike} I like eat {bones} and my birth year is {2016}//briefInfo是父类原型中的属性，并没有被继承，以下语句会报错dog1.briefInfo(); //error: dog1.briefInfo is not a functiondog2.briefInfo(); //error: dog2.briefInfo is not a function 存在的问题：
- 父类原型中定义的属性无法被继承
综合方式一方式二，一种很明显的方式呼之欲出了：
方式三：组合继承
实现方式：结合原型链继承和经典继承
子类
function Dog(name) { Animal.call(this, name); //经典继承}Dog.prototype = new Animal(); //原型链继承测试
var dog1 = new Dog('tom');dog1.food = 'shit';dog1.birthday.year = 2015;var dog2 = new Dog('mike');dog2.food = 'bones';dog2.birthday.year = 2016;dog1.greeting(); //console: Hi, my name is {tom} I like eat {shit} and my birth year is {2015}dog2.greeting(); //console: Hi, my name is {mike} I like eat {bones} and my birth year is {2016}dog1.briefInfo(); //console: {"name":"tom","food":"shit","birthday":{"year":2015}}dog2.briefInfo(); //console: {"name":"mike","food":"bones","birthday":{"year":2016}}//终于得到了预期的结果，算是较好的实现了继承。//但是，并不完美存在的问题
- 父类的构造函数被调用了两次
为了引出下一个继承方式，先将函数的继承放在一边，看一下js中对象的继承
(摘抄原文)
1. 原型式继承
var person = { name: 'martin' firend: ['bob', 'steven']};function object(o) { function F() {}; F.prototype = o; return new F();}var anotherPerson = object(person); //antherPerson继承了person2. 寄生式继承
function createAnother(original) { var clone = object(original); //原型式继承定义的方法 //扩展对象 clone.sayHi = function() { console.log('hi'); } return clone;}回到正题，接下来介绍一颗 “银弹”
方式四：寄生组合式继承
实现方式：
利用经典继承拷贝父类构造中的属性到子类
利用原型式继承创建一个继承父类原型的对象
将该对象的constructor属性指向子类的构造函数(寄生式继承:扩展对象)
将子类的prototype指向该对象
子类
function Dog(name) { Animal.call(this, name);}function F() {}var supProto = Animal.prototype;F.prototype = supProto;var subProto = new F();subProto.constructor = Dog;Dog.prototype = subProto;测试
var dog1 = new Dog('tom');dog1.food = 'shit';dog1.birthday.year = 2015;var dog2 = new Dog('mike');dog2.food = 'bones';dog2.birthday.year = 2016;dog1.greeting(); //console: Hi, my name is {tom} I like eat {shit} and my birth year is {2015}dog2.greeting(); //console: Hi, my name is {mike} I like eat {bones} and my birth year is {2016}dog1.briefInfo(); //console: {"name":"tom","food":"shit","birthday":{"year":2015}}dog2.briefInfo(); //console: {"name":"mike","food":"bones","birthday":{"year":2016}}关键字：JavaScript, 继承
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