Type erasure in C++: how boost::shared_ptr and boost::function work?

The idea is simple, you define a base class that has an interface with the functionality you need, and then inherit from it. Since the type erased class uses only that interface, the actual type underneath is forgotten and erased. Alternatively, if the only needed interface can be expressed as free functions you can store pointers to the free functions.

namespace detail {
   struct deleter_base {
      virtual ~deleter_base() {}
      virtual void operator()( void* ) = 0;
   };
   template <typename T>
   struct deleter : deleter_base {
      virtual void operator()( void* p ) {
         delete static_cast<T*>(p);
      }
   };
}
template <typename T>
class simple_ptr {
   T* ptr;
   detail::deleter_base* deleter;
public:
   template <typename U>
   simple_ptr( U* p ) {
      ptr = p;
      deleter = new detail::deleter<U>();
   }
   ~simple_ptr() {
      (*deleter)( ptr );
      delete deleter;
   }
};

This is a really simplified smart pointer, but the idea is there. In the particular case of shared_ptr, the deleter is stored as part of the reference count object, that is held by pointer.