title: Why must a substype assign a parent to to it's prototype outside the subtype's function?
body: <p>If I assign a parent to my subtype's prototype inside my function, my subtype's prototype ends up pointing to Subtype instead of SuperType. Why is this?</p>
  
  <pre><code>function SubType( name ){
  }               
  SubType.prototype = new SuperType(); 
  
   // THE ABOVE WORKS :)
  
   function SubType(){
          SubType.prototype = new SuperType(); 
   }   
  
  // THE ABOVE DOESN'T WORK :(
  </code></pre>