Javascript - 在构造函数或构造函数的原型(prototype)属性中设置属性？

Question

所以我听说过应该在构造函数的原型(prototype)属性中设置方法，这样它就不会有多个不同的实例。但是属性本身呢？哪个是最佳实践？如果是这样，构造函数不应该总是空的吗？

function Gadget(name, color) {
     this.name = name;
     this.color = color;
     this.whatAreYou = function(){
       return 'I am a ' + this.color + ' ' + this.name;
     }
}

这实际上应该是...？

function Gadget(name,color){}

Gadget.prototype.name = name;
Gadget.prototype.color = color;
Gadget.prototype.whatAreYou = function() {
   return 'I am a ' + this.color + ' ' + this.name;
};

Answer 1

如果在原型(prototype)上设置它，该属性将被所有实例共享。通常不是你想要的。我在 http://js-bits.blogspot.com/2014/10/understanding-prototypical-inheritance.html 上写了这篇博客.您通常希望每个小工具都有自己的名称和颜色。您将如何使用您建议的代码来做到这一点？

顺便说一句，你建议的代码有 undefined variable (名称、颜色)

通常的方法是在原型(prototype)上设置方法，在对象本身上设置常规值。除非您确实希望所有实例共享一个属性，否则这就像我们在静态类型语言中所说的静态属性。

举个例子

function Gadget(name,color){
    this.name = name;
    this.color = color;
    // Since all gadgets in on the prototype, this is shared by all instances;
    // It would more typically be attached to Gadget.allGadgets instead of the prototype
    this.allGadgets.push(this);
    // Note that the following would create a new array on the object itself
    // not the prototype
    // this.allGadgets = [];

}

Gadget.prototype.allGadgets = [];
Gadget.prototype.whatAreYou = function() {
    return 'I am a ' + this.color + ' ' + this.name;
};

要记住的一个重要概念是写入(赋值)总是应用于对象本身，而不是原型(prototype)。然而，读取将遍历原型(prototype)链寻找该属性。

也就是

function Obj() {
   this.map = {};
}

function SharedObj() {}
SharedObj.prototype.map = {};

var obj1 = new Obj();
var obj2 = new Obj();
var shared1 = new SharedObj();
var shared2 = new SharedObj();

obj1.map.newProp = 5;
obj2.map.newProp = 10;
console.log(obj1.map.newProp, obj2.map.newProp); // 5, 10

// Here you're modifying the same map
shared1.map.newProp = 5;
shared2.map.newProp = 10;
console.log(shared1.map.newProp, shared2.map.newProp); // 10, 10

// Here you're creating a new map and because you've written to the object directly    
// You don't have access to the shared map on the prototype anymore
shared1.map = {};
shared2.map = {};
shared1.map.newProp = 5;
shared1.map.newProp = 10;
console.log(shared1.map.newProp, shared2.map.newProp); // 5, 10