Why do std::shared_ptr<void> work

I found some code using std::shared_ptr to perform arbitrary cleanup at shutdown. At first I thought this code could not possibly work, but then I tried the following:

#include <memory>
#include <iostream>
#include <vector>

class test {
public:
  test() {
    std::cout << "Test created" << std::endl;
  }
  ~test() {
    std::cout << "Test destroyed" << std::endl;
  }
};

int main() {
  std::cout << "At begin of main.\ncreating std::vector<std::shared_ptr<void>>" 
            << std::endl;
  std::vector<std::shared_ptr<void>> v;
  {
    std::cout << "Creating test" << std::endl;
    v.push_back( std::shared_ptr<test>( new test() ) );
    std::cout << "Leaving scope" << std::endl;
  }
  std::cout << "Leaving main" << std::endl;
  return 0;
}

This program gives the output:

At begin of main.
creating std::vector<std::shared_ptr<void>>
Creating test
Test created
Leaving scope
Leaving main
Test destroyed

I have some ideas on why this might work, that have to do with the internals of std::shared_ptrs as implemented for G++. Since these objects wrap the internal pointer together with the counter the cast from std::shared_ptr<test> to std::shared_ptr<void> is probably not hindering the call of the destructor. Is this assumption correct?

And of course the much more important question: Is this guaranteed to work by the standard, or might further changes to the internals of std::shared_ptr, other implementations actually break this code?

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

Why do std::shared_ptr<void> work

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