Why does TypeScript add "prototype"?

Question

My typescript code:

export class File {
    isOpenEnabled() {
        return false;
    }

    openClicked() {
        debugger;
    }
}

define([], function () {
    return {
        handler: new File()
    };
});

Is turned into:

define(["require", "exports"], function(require, exports) {
    var File = (function () {
        function File() {
        }
        File.prototype.isOpenEnabled = function () {
            return false;
        };

        File.prototype.openClicked = function () {
            debugger;
        };
        return File;
    })();
    exports.File = File;

    define([], function () {
        return {
            handler: new File()
        };
    });
});

Why the insertion of prototype?

thanks - dave

Alberto · Accepted Answer · 2013-11-09 21:17:03Z

2

Functions in javascript are objects.

For example:

function MyClass () {

  this.MyMethod= function () {};
}

Every time a new instance of MyClass is created, a new instance of MyMethod is also created. A better approach is to add the function MyMethod to the prototype of MyClass:

MyClass.prototype.MyMethod = function(){};

In this way no matter how many instances of MyClass you create, only a single MyMethod will be created.

Back to your question I think that typescript is doing exactly this kind of optimization for the methods that you defined in your File class.

answered Nov 9, 2013 at 21:17

Alberto

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1 Comment

David Thielen Over a year ago

So it creates MyClass.prototype once and it is then assigned to all instances of the MyClass object? Thanks - dave

basarat · Accepted Answer · 2013-11-10 23:47:00Z

1

There are two reason:

Memory optimization (already mentioned by Alberto)
Prototypical Inheritance

The second reason as well as the first is well covered in : http://javascript.crockford.com/inheritance.html

answered Nov 10, 2013 at 23:47

basarat

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