在多级继承中查找构造函数

Question

I want to find out the specific constructor used to instantiate the object in javascript, not the one which is last in the prototype chain. Consider the code :

function F(){};
function E(){};
function D(){};
function C(){};
function B(){};
function A(){};

E.prototype= new F();
D.prototype= new E();
C.prototype= new D();
B.prototype= new C();
A.prototype= new B();

a=new A();

Find fiddle here

a.constructor returns function F(){} , but I want a method that returns function A(){} , since A is the constructor used to instantiate the object.

How can that be achieved ?

Answer 1

With the way you inherit from parent it's not possible to access original constructor, because when you write

A.prototype = new B();

A.prototype.constructor indeed points to B not A anymore.

With this pattern for prototypical inheritance you have to manually set constructor properly. So you either do it for each extended class manually or you can use helper function:

 function inherit(C, P) { C.prototype = new P(); C.prototype.constructor = C; } function F(){}; function E(){}; function D(){}; function C(){}; function B(){}; function A(){}; inherit(E, F); inherit(D, E); inherit(C, D); inherit(B, C); inherit(A, B); var a = new A(); var c = new C() document.write( a.constructor + "<br>" ); document.write( c.constructor ); 

Answer 2

Do the following:

function F(){};
function E(){};
function D(){};
function C(){};
function B(){};
function A(){};

E.prototype= new F();
E.prototype.constructor = E;
D.prototype= new E();
D.prototype.constructor = D;
C.prototype= new D();
C.prototype.constructor = C;
B.prototype= new C();
B.prototype.constructor = B;
A.prototype= new B();
A.prototype.constructor = A;

Answer 3

Did you check John Resig's Class.js ? http://ejohn.org/blog/simple-javascript-inheritance/ It implements inheritance in javascript, and makes possible what you're asking for.

Edit: Sorry. I was wrong. However, you can add something like this to your declarations :

A.prototype= new B();
A.prototype.constructor = B; 

Not sure of all the consequences it can have, but it seems to work!