Switching out an object's prototype

Question

I'm trying to understand how prototypes work. Why does the following break?

var A = function A(){this.a = 0},
    aa = new A;

A.prototype = {hi:"hello"};

aa.constructor.prototype //->{hi:"hello"} ok so far :)

aa.hi //undefined?? why? :(

Answer 1

I think you meant in the last line aa.hi instead of aa.hello.

It gives you undefined because the A.prototype is assigned after the new object (aa) has been already created.

In your second line:

//...
aa = new A;
//...

This will create an object that inherits from A.prototype, at this moment, A.prototype is a simple empty object, that inherits from Object.prototype.

This object will remain referenced by the internal [[Prototype]] property of the aa object instance.

Changing A.prototype after this, will not change the direct inheritance relationship between aa and that object.

In fact, there is no standard way to change the [[Prototype]] internal property, some implementations give you access through a non-standard property called __proto__.

To get the expected results, try:

var A = function A () { this.a = 0 };
A.prototype = { hi:"hello" };

var aa = new A;

aa.hi; // "hello"