Why do std::shared_ptr<void> work

Question

The trick is that std::shared_ptr performs type erasure. Basically, when a new shared_ptr is created it will store internally a deleter function (which can be given as argument to the constructor but if not present defaults to calling delete). When the shared_ptr is destroyed, it calls that stored function and that will call the deleter.

A simple sketch of the type erasure that is going on simplified with std::function, and avoiding all reference counting and other issues can be seen here:

template <typename T>
void delete_deleter( void * p ) {
   delete static_cast<T*>(p);
}

template <typename T>
class my_unique_ptr {
  std::function< void (void*) > deleter;
  T * p;
  template <typename U>
  my_unique_ptr( U * p, std::function< void(void*) > deleter = &delete_deleter<U> ) 
     : p(p), deleter(deleter) 
  {}
  ~my_unique_ptr() {
     deleter( p );   
  }
};

int main() {
   my_unique_ptr<void> p( new double ); // deleter == &delete_deleter<double>
}
// ~my_unique_ptr calls delete_deleter<double>(p)

When a shared_ptr is copied (or default constructed) from another the deleter is passed around, so that when you construct a shared_ptr<T> from a shared_ptr<U> the information on what destructor to call is also passed around in the deleter.

shared_ptr<T> logically[*] has (at least) two relevant data members:

• a pointer to the object being managed
• a pointer to the deleter function that will be used to destroy it.

The deleter function of your shared_ptr<Test>, given the way you constructed it, is the normal one for Test, which converts the pointer to Test* and deletes it.

When you push your shared_ptr<Test> into the vector of shared_ptr<void>, both of those are copied, although the first one is converted to void*.

So, when the vector element is destroyed taking the last reference with it, it passes the pointer to a deleter that destroys it correctly.

It's actually a little more complicated than this, because shared_ptr can take a deleter functor rather than just a function, so there might even be per-object data to be stored rather than just a function pointer. But for this case there is no such extra data, it would be sufficient just to store a pointer to an instantiation of a template function, with a template parameter that captures the type through which the pointer must be deleted.

[*] logically in the sense that it has access to them - they may not be members of the shared_ptr itself but instead of some management node that it points to.

It works because it uses type erasure.

Basically, when you build a shared_ptr, it passes one extra argument (that you can actually provide if you wish), which is the deleter functor.

This default functor accepts as argument a pointer to type you use in the shared_ptr, thus void here, casts it appropriately to the static type you used test here, and calls the destructor on this object.

Any sufficiently advanced science feels like magic, isn't it ?

The constructor shared_ptr<T>(Y *p) indeed seems to be calling shared_ptr<T>(Y *p, D d) where d is an automatically generated deleter for the object.

When this happens the type of the object Y is known, so the deleter for this shared_ptr object knows which destructor to call and this information is not lost when the pointer is the stored in a vector of shared_ptr<void>.

Indeed the specs require that for a receving shared_ptr<T> object to accept a shared_ptr<U> object it must be true that and U* must be implicitly convertible to a T* and this is certainly the case with T=void because any pointer can be converted to a void* implicitly. Nothing is said about the deleter that will be invalid so indeed the specs are mandating that this will work correctly.

Technically IIRC a shared_ptr<T> holds a pointer to an hidden object that contains the reference counter and a pointer to the actual object; by storing the deleter in this hidden structure it's possible to make this apparently magic feature working while still keeping shared_ptr<T> as big as a regular pointer (however dereferencing the pointer requires a double indirection

shared_ptr -> hidden_refcounted_object -> real_object

Test* is implicitly convertible to void*, therefore shared_ptr<Test> is implicitly convertible to shared_ptr<void>, from memory. This works because shared_ptr is designed to control destruction at run-time, not compile-time, they will internally use inheritance to call the appropriate destructor as it was at allocation time.

I am going to answer this question (2 years later) using a very simplistic implementation of shared_ptr that the user will understand.

Firstly I am going to a few side classes, shared_ptr_base, sp_counted_base sp_counted_impl, and checked_deleter the last of which is a template.

class sp_counted_base
{
 public:
    sp_counted_base() : refCount( 1 )
    {
    }

    virtual ~sp_deleter_base() {};
    virtual void destruct() = 0;

    void incref(); // increases reference count
    void decref(); // decreases refCount atomically and calls destruct if it hits zero

 private:
    long refCount; // in a real implementation use an atomic int
};

template< typename T > class sp_counted_impl : public sp_counted_base
{
 public:
   typedef function< void( T* ) > func_type;
    void destruct() 
    { 
       func(ptr); // or is it (*func)(ptr); ?
       delete this; // self-destructs after destroying its pointer
    }
   template< typename F >
   sp_counted_impl( T* t, F f ) :
       ptr( t ), func( f )

 private:

   T* ptr; 
   func_type func;
};

template< typename T > struct checked_deleter
{
  public:
    template< typename T > operator()( T* t )
    {
       size_t z = sizeof( T );
       delete t;
   }
};

class shared_ptr_base
{
private:
     sp_counted_base * counter;

protected:
     shared_ptr_base() : counter( 0 ) {}

     explicit shared_ptr_base( sp_counter_base * c ) : counter( c ) {}

     ~shared_ptr_base()
     {
        if( counter )
          counter->decref();
     }

     shared_ptr_base( shared_ptr_base const& other )
         : counter( other.counter )
     {
        if( counter )
            counter->addref();
     }

     shared_ptr_base& operator=( shared_ptr_base& const other )
     {
         shared_ptr_base temp( other );
         std::swap( counter, temp.counter );
     }

     // other methods such as reset
};

Now I am going to create two "free" function called make_sp_counted_impl which will return a pointer to a newly created one.

template< typename T, typename F >
sp_counted_impl<T> * make_sp_counted_impl( T* ptr, F func )
{
    try
    {
       return new sp_counted_impl( ptr, func );
    }
    catch( ... ) // in case the new above fails
    {
        func( ptr ); // we have to clean up the pointer now and rethrow
        throw;
    }
}

template< typename T > 
sp_counted_impl<T> * make_sp_counted_impl( T* ptr )
{
     return make_sp_counted_impl( ptr, checked_deleter<T>() );
}

Ok, these two functions are essential as to what will happen next when you create a shared_ptr through a templated function.

template< typename T >
class shared_ptr : public shared_ptr_base
{

 public:
   template < typename U >
   explicit shared_ptr( U * ptr ) :
         shared_ptr_base( make_sp_counted_impl( ptr ) )
   {
   }

  // implement the rest of shared_ptr, e.g. operator*, operator->
};

Note what happens above if T is void and U is your "test" class. It will call make_sp_counted_impl() with a pointer to U, not a pointer to T. The management of the destruction is all done through here. The shared_ptr_base class manages the reference counting with regards to copying and assignment etc. The shared_ptr class itself manages the typesafe use of operator overloads (->, * etc).

Thus although you have a shared_ptr to void, underneath you are managing a pointer of the type you passed into new. Note that if you convert your pointer to a void* before putting it into the shared_ptr, it will fail to compile on the checked_delete so you are actually safe there too.