1. 原型鏈繼承，缺點：引用類型屬性list改了，會影響其他實例

function Super() {
  this.list = [1,2]
}

Super.prototype.sayAge = function () {
  console.log(this.age)
}

function Sub(age) {
  this.age = age;
}

Sub.prototype = new Super()

var s1 = new Sub(30) 
s1.list.push(3)
var s2 = new Sub(30) 
s2.list // [1,2,3]

2. 借用構造函數(shù)，函數(shù)被call來使用，無法繼承函數(shù)的原型

function Super() {
  this.list = [1,2]
}

Super.prototype.sayAge = function () {
  console.log(this.age)
}

function Sub(age) {
  this.age = age;
  Super.call(this)
} 

var s1 = new Sub(30)
s1.sayAge // undefined

3. 組合繼承；結合上面的兩個繼承，互補利弊；唯一不足，實例通過Super.call有的屬性，原型上 new Super()也有，重復了

function Super(name) {
  this.name = name
  this.list = [1,2]
}

Super.prototype.sayAge = function () {
  console.log(this.age)
}

function Sub(name, age) {
  Super.call(this, name)
  this.age = age;
} 

Sub.prototype = new Super()

var s1 = new Sub('kp', 30)

s1原型上，屬性重復

4. 寄生組合式繼承，Sub.prototype是一個空對象new F()【代替new Super()】這個空對象的構造函數(shù)原型是Super.prototype；解決上面問題： 原型上 new Super()也有，重復了 的問題

function object(o){ 
  function F(){} 
  F.prototype = o; 
  return new F(); 
} 

function Super(name) {
  this.name = name
  this.list = [1,2]
}

Super.prototype.sayAge = function () {
  console.log(this.age)
}

function Sub(name, age) {
  Super.call(this, name)
  this.age = age;
} 

Sub.prototype = object(Super.prototype)
// Sub.prototype = Object.create(Super.prototype) // 上一行代碼，可以用Object.create，代替object這個自定義方法
Sub.prototype.constructor = Sub
var s1 = new Sub('kp', 30)
s1

s1原型上，沒有重復屬性

5. class 繼承，采用的就是 【4. 寄生組合式繼承】

class Super{
  constructor(name) {
    this.name = name
  }

  sayAge() {
    console.log(this.age)
  }
}

class Sub extends Super{
  constructor(name, age) {
    super(name)
    this.age = age
  }
}

var s1 = new Sub('kp', 30)

s1 - class繼承