对 javascript 的构造函数和原型感到困惑？

Question

function MyObject(){}
Array.prototype={};
MyObject.prototype={};
var a=new Array();
var b=new MyObject();
alert(a.constructor==Array);//true
alert(b.constructor==MyObject);//false

Answer 1

Array.prototype is a non-writable property.

As such, your assignment:

Array.prototype = {}

...doesn't succeed, and so its .constructor property hasn't changed.

15.4.3.1 Array.prototype

The initial value of Array.prototype is the Array prototype object (15.4.4).

This property has the attributes { [[Writable]]: false, [[Enumerable]]: false, [[Configurable]]: false }

...whereas with your custom constructor, you have the ability to assign a different prototype object, so you've overwritten the original which had reference to the constructor via .constructor .

Answer 2

The constructor property is overwritten when you override the prototype property with your own empty object instance, since ({}).constructor === Object . You can do either

function MyObject() {}
MyObject.prototype = {};
MyObject.prototype.constructor = MyObject;

or (better IMO) you can not set prototype directly, but instead augment it:

function MyObject() {}
MyObject.prototype.foo = "bar";

Also of note: Array.prototype is not writable, so your line Array.prototype = {} will silently fail (or noisily fail in strict mode).

Answer 3

> function MyObject(){} 
> Array.prototype={};

You can't assign a value to Array.prototype.

> MyObject.prototype={};
> var a=new Array();
> var b=new MyObject();
> alert(a.constructor==Array);//true

Array.prototype has a constructor property that references the Array function. Since a is an instance of Array , it inherits Array.prototype 's constructor property.

> alert(b.constructor==MyObject);//false

You have assigned an empty object to MyObject.prototype , it does not have a prototype property, nor does b .

MyObject.prototype.constructor = MyObject;

alert(b.constructor==MyObject); // true