javascript - Switching out an object's prototype - Stack Overflow

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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? :(

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? :(

• javascript

Share Improve this question Follow edited May 21, 2015 at 3:54 hello_there_andy 2,08222 gold badges2222 silver badges5252 bronze badges asked Dec 30, 2010 at 18:35 JohnJohn 19711 silver badge88 bronze badges 2

• 1 woops, made a correction to aa.hi – John Commented Dec 30, 2010 at 18:48
• Removed salutation: "Thanks in advance!", don't do it next time – hello_there_andy Commented May 21, 2015 at 2:38

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1 Answer 1

Sorted by: Reset to default 12

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"

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