title: What is the best way to implement smart pointers in C++?
body: <p>I've been evaluating various smart pointer implementations (wow, there are a LOT out there) and it seems to me that most of them can be categorized into two broad classifications:</p>
  
  <p>1) This category uses inheritance on the objects referenced so that they have reference counts and usually up() and down() (or their equivalents) implemented. IE, to use the smart pointer, the objects you're pointing at must inherit from some class the ref implementation provides.</p>
  
  <p>2) This category uses a secondary object to hold the reference counts. For example, instead of pointing the smart pointer right at an object, it actually points at this meta data object... Who has a reference count and up() and down() implementations (and who usually provides a mechanism for the pointer to get at the actual object being pointed to, so that the smart pointer can properly implement operator ->()).</p>
  
  <p>Now, 1 has the downside that it forces all of the objects you'd like to reference count to inherit from a common ancestor, and this means that you cannot use this to reference count objects that you don't have control over the source code to.</p>
  
  <p>2 has the problem that since the count is stored in another object, if you ever have a situation that a pointer to an existing reference counted object is being converted into a reference, you probably have a bug (I.E., since the count is not in the actual object, there is no way for the new reference to get the count... ref to ref copy construction or assignment is fine, because they can share the count object, but if you ever have to convert from a pointer, you're totally hosed)...</p>
  
  <p>Now, as I understand it, boost::shared_pointer uses mechanism 2, or something like it... That said, I can't quite make up my mind which is worse! I have only ever used mechanism 1, in production code... Does anyone have experience with both styles? Or perhaps there is another way thats better than both of these?</p>