了解Javascript的OOP:修改其他实例的实例 [英] Understanding Javascript's OOP: instances modifying other instances

问题描述

我很难理解为什么修改实例a中的属性会修改实例b上的相同属性.

I'm having some trouble understanding why modifying a property in instance a modifies the same property on instance b.

var A = function (){

};

A.prototype.data = {
    value : 0
};

var a = new A();
var b = new A();

console.log(a.data.value, b.data.value); // 0, 0
a.data.value = 5;
console.log(a.data.value, b.data.value); // 5, 5

prototype关键字不应该将data变量设置为实例变量吗?

Shouldn't the prototype keyword make the data variable an instance variable ?

在按照预期执行的示例中，情况似乎并非如此:

This seems not to be the case in this example which executes as expected:

var B = function (){
    this.data = {
        value : 0
    };
};

var i = new B();
var j = new B();

console.log(i.data.value, j.data.value); // 0, 0
i.data.value = 5;
console.log(i.data.value, j.data.value); // 5, 0

我很困惑为什么原型方法不起作用.也许我缺少一些概念性知识.

I'm confused as why the prototype method wont work. Maybe i'm lacking some conceptual knowledge.

推荐答案

js中的原型是该类所有实例之间的数据共享，就像经典oop语言中的静态变量一样.有时在.prototype中放置一些方法/字段以节省运行时的内存很有用.

Prototype in js is a data shared between all instances of that class, it's like a static variable from classic oop languages. Sometimes it's usefull to put some methods/fields in .prototype to save memory on runtime.

这篇关于了解Javascript的OOP:修改其他实例的实例的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！