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    <p>It looks similar to the Crockford's <code>Object.create</code> method, but this function is used to "setup" constructors.</p> <p>It accepts two constructors as arguments, and it setups the <code>prototype</code> of the first one.</p> <p>Let me rename the cryptic variable names:</p> <pre><code>function G(sub, super) { var F = function() { }; F.prototype = super.prototype; sub.superLink = super.prototype; sub.prototype = new F(); } function Super () { //... } Super.prototype.member1 = 'superMember1'; function Sub() { this.member2 = 'subMember2'; } G(Sub, Super); new Sub(); // Object { member2="subMember2", member1="superMember1"} </code></pre> <p><strong>Edit:</strong> The <code>T</code> property is simply used to know what is the "super" constructor of the sub one, I've seen this pattern on other places, like in the book <em>Pro JavaScript Design Patterns</em> (<a href="http://books.google.com/books?id=za3vlnlWxb0C&amp;lpg=PP1&amp;dq=Pro%20JavaScript%20Design%20Patterns&amp;pg=PA43#v=onepage&amp;q&amp;f=false" rel="nofollow noreferrer">page 43</a>), with some additions, to prevent the <code>constructor</code> property to point to the wrong object:</p> <pre><code>function extend(subClass, superClass) { var F = function() {}; F.prototype = superClass.prototype; subClass.prototype = new F(); subClass.prototype.constructor = subClass; subClass.superclass = superClass.prototype; if(superClass.prototype.constructor == Object.prototype.constructor) { superClass.prototype.constructor = superClass; } } </code></pre> <p>See also: </p> <ul> <li><a href="https://stackoverflow.com/questions/2686258/javascript-inheritance-extend-function">JavaScript inheritance extend function</a></li> </ul>
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    1. COI don't think its that simple, the `T` property is used in some way not clear by the question. Also, the result from your answer can be reached by just setting `Sub.prototype = Super.prototype`
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      1. USSean Kinsey
    2. CO@Sean Not quite. By doing that, if you add a member to `Sub.prototype` you will also add a member to `Super.prototype`.
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      1. USPablo Cabrera
    3. CO@CMS I think it is worth to note that `member2` will be on the prototype chain of `new Sub()` and not on `new Sub()` itself.
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