以下内容均基于本人对《JavaScript高等程序设计》第三版6.3小节的明白

先看一下父类

function Animal(name) {
    var name = name;             //'私有（受庇护）'成员，只允许在父类的组织函数中赋值
    this.food = undefined;       //'公有'成员

    //援用范例的成员
    this.birthday = {            
      year: undefined
    };

    //组织函数中的要领，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 + '}');
    };

    //原型中的要领，将组织函数中的非函数成员以JSON花样打印
    Animal.prototype.briefInfo = function() {
      var brief = {
        name: name,
        food: this.food,
        birthday: this.birthday
      };
      console.log(JSON.stringify(brief));
    };
  }

体式格局一：原型链继续

完成体式格局：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是援用范例的属性，所以dog1的birthday.year被dog2覆蓋了;
//没法给dog1和dog2的name赋值

存在的题目：

• 援用范例的对象会被子类的一切实例同享(1.1)

• 没法在建立子类的实例时，给父类的组织函数通报参数(1.2)

体式格局二：借用组织函数（捏造对象、典范继续）

完成体式格局：在子类的组织函数中应用call(或许apply)要领实行父类组织函数(题目1.2处理)，将实行对象设为子类的this,相当于把父类组织函数中的成员拷贝了一份到子类(题目1.1处理)

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，并没有被继续，以下语句会报错
dog1.briefInfo();  //error: dog1.briefInfo is not a function
dog2.briefInfo();  //error: dog2.briefInfo is not a function 

存在的题目：

1. 父类原型中定义的属性没法被继续

综合体式格局一体式格局二，一种很明显的体式格局呼之欲出了：

体式格局三：组合继续

完成体式格局：连系原型链继续和典范继续

子类

function Dog(name) {
  Animal.call(this, name);    //典范继续
}
Dog.prototype = new Animal();    //原型链继续

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}}

//终究得到了预期的效果，算是较好的完成了继续。
//然则，并不圆满

存在的题目

• 父类的组织函数被调用了两次

为了引出下一个继续体式格局，先将函数的继续放在一边，看一下js中对象的继续

(摘抄原文)

1. 原型式继续

var person = {
  name: 'martin'
  firend: ['bob', 'steven']
};
  
function object(o) {
  function F() {};
  F.prototype = o;
  return new F();
}
var anotherPerson = object(person);    //antherPerson继续了person

2. 寄生式继续

function createAnother(original) {
  var clone = object(original);    //原型式继续定义的要领
  //扩大对象
  clone.sayHi = function() {
    console.log('hi');
  }
  return clone;
}

回到正题，接下来引见一颗 “银弹”

体式格局四：寄生组合式继续

完成体式格局：

1. 应用典范继续拷贝父类组织中的属性到子类

2. 应用原型式继续建立一个继续父类原型的对象

3. 将该对象的constructor属性指向子类的组织函数(寄生式继续:扩大对象)

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}}