[boost] enable_shared_from_this extension

From: Raffaele Romito (berserker_r_at_[hidden])
Date: 2006-01-10 10:58:23

I often need to retrive a shared_ptr of a derived that inherit from a 
"shared_enabled" base class, for example:

class A : public boost::enable_shared_from_this<A>
{
public:
   A() { }
   virtual ~A() { }

   void some_a_method()
   {
      boost::shared_ptr<A> ptr = shared_from_this();
   }
};

class B : public A
{
public:
   void some_b_method()
   {
      boost::shared_ptr<B> ptr = 
boost::dynamic_pointer_cast<B>(shared_from_this());
   }
};

Actually I'm using an helper class to "hide" the dynamic_pointer_class 
stuff:

template <typename T>
class shared_enabled : public boost::enable_shared_from_this<T>
{
public:
      shared_enabled() { }
      virtual ~shared_enabled() { }

      inline boost::shared_ptr<T> this_ptr() { return shared_from_this(); }

      inline boost::shared_ptr<T const> this_ptr() const { return 
shared_from_this(); }

      template <typename I>
      inline boost::shared_ptr<I> this_ptr() { return 
boost::dynamic_pointer_cast<I>(shared_from_this()); }

      template <typename I>
      inline boost::shared_ptr<I const> this_ptr() const { return 
boost::dynamic_pointer_cast<I const>(shared_from_this()); }
};

Deriving A from shared_enabled it's easier (my opinion) to write 
"some_b_method":

void some_b_method()
{
      boost::shared_ptr<B> ptr = this_ptr<B>();
}

What about a simple extension in the enable_shared_from_this class?

template <typename I>
inline shared_ptr<I> shared_from_this()
{
        return dynamic_pointer_cast<I>(shared_from_this());
}

template <typename I>
inline shared_ptr<I const> shared_from_this() const
{
        return dynamic_pointer_cast<I const>(shared_from_this());
}