关于Javascript中通过实例对象修改原型对象属性值的问题

function Animal() {}
Animal.prototype = {
 constructor: Animal,
number: "very much",
 fish: ["shark","sardine"],
 bird:{
     ability: "fly",
     feature: "feather"
 }
};
var animal1 = new Animal();
var animal2 = new Animal();
//没有改变原型的number属性,而是自己获得了number属性
animal1.number = 1000;
console.log(animal2.number);//very much

function Animal() {}
Animal.prototype = {
 constructor: Animal,
 number: "very much",
 fish: ["shark","sardine"],
 bird:{
     ability: "fly",
     feature: "feather"
 }
};
var animal1 = new Animal();
var animal2 = new Animal();
var animal3 = new Animal();
//没有改写原型中的fish属性,此时animal1实例对象中有了自己的fish属性，向其自己的fish属性中推入和弹出项不会改变原型的fish属性。
animal1.fish = ["cold fish"];
for (var i=0;i<animal2.fish.length;i++) {
  console.log(animal2.fish[i]);//shark,sardine,没有cold fish
 }
//通过实例对象animal3向fish属性中推入项，改变了原型对象的fish属性，因为实例对象中没有自己的fish属性
animal3.fish.push("voladao");
animal3.fish[0] = "fly fish";
for (var i=0;i<animal2.fish.length;i++) {
  console.log(animal2.fish[i]);//fly fish,sardine,voladao
 }

function Animal() {}
Animal.prototype = {
 constructor: Animal,
 number: "very much",
 fish: ["shark","sardine"],
 bird:{
     ability: "fly",
     feature: "feather"
 }
}
var animal1 = new Animal();
var animal2 = new Animal();
var animal3 = new Animal();
var animal4 = new Animal();
var animal5 = new Animal();
//改写了原型中的bird属性
animal1.bird.ability = "run";
console.log(animal2.bird.ability);//run
//创建animal3的bird属性，没有改变原型对象的bird属性
 animal3.bird = {
   eat: "fish"
 };
 console.log(animal4.bird.eat);//undifined
 console.log(animal3.bird.eat);//fish
 animal5.bird.home = "tree";
 console.log(animal4.bird.home);//tree