(1)
function Aa(name){
this.name = name;
this.colors = ["red","blue","green"];
}
Aa.prototype.sayName = function(){
console.log(this.name)
}
function Bb(name,age){
Aa.call(this,name);
this.age = age;
}
var b = new Bb()
// console.log(b.sayName)
console.log(Aa.prototype)
Bb.prototype.sayAge = function (){
console.log(this.age)
}
Bb.prototype.ue="ni"
console.log(Bb.prototype)
未重写Bb.prototype原型前，Aa.prototype原型所包含的属性方法有一下这些：
未重写Bb.prototype原型前，Bb.prototype原型所包含的属性方法有一下这些：
同样这里，实例b也不能访问Aa的原型方法和属性
(2)
function Aa(name){
this.name = name;
this.colors = ["red","blue","green"];
}
Aa.prototype.sayName = function(){
console.log(this.name)
}
function Bb(name,age){
Aa.call(this,name);
this.age = age;
}
var b = new Bb()
// console.log(b.sayName)
console.log(Aa.prototype)
Bb.prototype.sayAge = function (){
console.log(this.age)
}
Bb.prototype.ue="ni"
console.log(Bb.prototype)
//重写Bb.prototype
Bb.prototype = new Aa();
Bb.prototype.constructor = Bb;
console.log(Bb.prototype)
console.log(instance.ue)//undefined
instance.sayAge()//直接报错，说instance.sayAge is not a function
重写Bb.prototype后，Bb.prototype的属性和方法：将Aa构造函数的属性和原型的方法都继承过来了，但是替换前添加的Bb原型对象的方法和属性都将被重写，所以当Bb的实例instance访问Bb的原型属性ue是会打印出undefined
当访问sayAge是，会报错instance.sayAge is not a function
总结：为了避免超类型构造函数不会重写子类型的属性，可以在调用超类型构造函数后，再添加应该再子类型中定义的属性和方法