JavaScript实现继承

如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，只容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，因此dog1的birthday.year被dog2覆盖了;
//没法给dog1和dog2的name赋值

存在的问题：

• 引用类型的对象会被子类的全部实例共享(1.1)

• 没法在建立子类的实例时，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,至关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，并无被继承，如下语句会报错
dog1.briefInfo();  //error: dog1.briefInfo is not a function
dog2.briefInfo();  //error: dog2.briefInfo is not a function

存在的问题：

1. 父类原型中定义的属性没法被继承

---

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}}

//终于获得了预期的结果，算是较好的实现了继承。
//可是，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person

2. 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：

1. 利用经典继承拷贝父类构造中的属性到子类

2. 利用原型式继承建立一个继承父类原型的对象

3. 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)

4. 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}}