a question about inheritance of Array in javascript

Question

function ClassA()  
{  
    this.a=[];
    this.aa=100;  
}  


function ClassB()  
{  
    this.b=function(){return "classbb"};  
}  
ClassB.prototype=new ClassA();  
Array.prototype= new ClassB();  
var array1= new Array();
alert(array1.b());

Why can't Array inherit ClassA and ClassB? Thanks.

I hope we both agree that this question is of theoretical nature, since Array would be rendered useless if you overwrote its prototype. — user123444555621
– user123444555621, Commented Mar 25, 2011 at 8:04

KooiInc · Accepted Answer · 2011-03-25 08:13:31Z

3

That's not quite the way to make the Array.prototype inherit from your objects. It would overwrite Array.prototype, which isn't allowed obviously.

You can however extend the prototype of Array with the properties/methods of ClassA/ClassB like this:

function ClassA() {  
  this.a=[];
  this.aa=100;  
}  

function ClassB() {  
  this.b=function(){return "classbb"};  
}

ClassB.prototype = new ClassA; 

var instB = new ClassB;
for (var l in instB){
    Array.prototype[l] = instB[l];
}

var array1 = [];
alert(array1.aa);

You can also:

Array.prototype.classb = new ClassB;
var array1 = [];
alert(array1.classb.aa);

edited Mar 25, 2011 at 8:13

answered Mar 25, 2011 at 8:01

KooiInc

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6 Comments

jsnewman Over a year ago

Can I write it this way? function ClassA() { this.a=[]; this.aa=100; } function ClassB() { this.b=function(){return "classbb"}; } ClassB.prototype = new ClassA; for (var l in ClassB.prototype){ Array.prototype[l] = ClassB.[l]; } var array1 = []; alert(array1.b);

KooiInc Over a year ago

Not really. First, you'll need to assign Classb.prototype[l] (which you are enumerating in a for loop). Second, if you do so, you will miss ClassB.b, because that's not in ClassB's prototype. It would work if you added method b to ClassB's prototype. See jsfiddle.net/pgw9q

jsnewman Over a year ago

Could you tell me why it still doesn't work if I write it this way:function ClassA() { this.a=[]; this.aa=100; } function ClassB() { } ClassB.prototype = new ClassA; ClassB.prototype.b=function(){return "classbb"}; for (var l in ClassB.prototype){ // console.log(l); Array.prototype[l] = ClassB.[l]; //I delete prototype here } var array1 = []; alert(array1.b());

KooiInc Over a year ago

ClassB.[l] should be ClassB.prototype[l]. What do you mean with "I delete prototype here"? Nothing is deleted in your code.

jsnewman Over a year ago

Why can't ClassB.prototype[l] be replaced by ClassB[l]? I am new in javascript. Please forgive my ignorance.

|

user123444555621 · Accepted Answer · 2011-03-25 08:00:04Z

1

The standard prohibits overwriting 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 }.

You can easily verify that browsers comply to this:

var origArrayProto = Array.prototype;
Array.prototype = new function () {}; // try to overwrite
alert(Array.prototype == origArrayProto); // true

answered Mar 25, 2011 at 8:00

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