- 原型链继承，缺点：引用类型属性list改了，会影响其他实例
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]
- 借用构造函数，函数被call来使用，无法继承函数的原型
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
- 组合继承；结合上面的两个继承，互补利弊；唯一不足，实例通过Super.call有的属性，原型上 new Super()也有，重复了
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，属性重复
- 寄生组合式继承，Sub.prototype是一个空对象new F()【代替new Super()】这个空对象的构造函数原型是Super.prototype；解决上面问题： 原型上 new Super()也有，重复了 的问题
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) // 上一行代码，可以用Object.create，代替object这个自定义方法
Sub.prototype.constructor = Sub
var s1 = new Sub('kp', 30)
s1
s1原型上，没有重复属性
- class 继承，采用的就是 【4. 寄生组合式继承】
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继承
