Shared void pointers. Why does this work?

Question

To solve a very peculiar problem in my application I need a shared-pointer to allocated data, but to the outside world, the underlying data type should remain hidden.

I could solve this by making some kind of Root class of which all my other classes inherit, and use a shared_ptr on this Root class, like this:

std::shared_ptr<Root>

However:

• I don't want all my classes to inherit from this Root class just to be able to have this shared pointer
• Sometimes I want to return a shared pointer to std::vector, or std::list, or std::set, ... which obviously don't inherit from my Root class

Strange enough, it seems that you can create a shared_ptr on void and this seems to work correctly, like shown in this example:

class X
   {
   public:
      X() {std::cout << "X::ctor" << std::endl;}
      virtual ~X() {std::cout << "X::dtor" << std::endl;}
   };

typedef std::shared_ptr<void> SharedVoidPointer;

int main()
{
X *x = new X();
SharedVoidPointer sp1(x);
}

x is correctly deleted and in a larger experiment I could verify that the shared pointer does indeed what it needs to do (delete x afer the last shared_ptr turns out the light).

Of course this solves my problem, since I can now return data with a SharedVoidPointer data member and be sure that it's correctly cleaned up where it should be.

But is this guaranteed to work in all cases? It clearly works in Visual Studio 2010, but does this also work correctly on other compilers? On other platforms?

Answer 1

The shared_ptr constructor that you use is actually a constructor template that looks like:

template <typename U>
shared_ptr(U* p) { }

It knows inside of the constructor what the actual type of the pointer is (X) and uses this information to create a functor that can correctly delete the pointer and ensure the correct destructor is called. This functor (called the shared_ptr's "deleter") is usually stored alongside the reference counts used to maintain shared ownership of the object.

Note that this only works if you pass a pointer of the correct type to the shared_ptr constructor. If you had instead said:

SharedVoidPointer sp1(static_cast<void*>(x));

then this would not have worked because in the constructor template, U would be void, not X. The behavior would then have been undefined, because you aren't allowed to call delete with a void pointer.

In general, you are safe if you always call new in the construction of a shared_ptr and don't separate the creation of the object (the new) from the taking of ownership of the object (the creation of the shared_ptr).

Answer 2

I think the implicit point of the question was that you can't delete by void*, so it seems strange that you can delete through shared_ptr<void>.

You can't delete an object via a raw void* primarily because it wouldn't know what destructor to call. Using a virtual destructor doesn't help because void doesn't have a vtable (and thus no virtual destructor).

James McNellis clearly explained why shared_ptr<void> works, but there is something else interesting here: Assuming you follow the documented best practice to always use the following form when invoking new...

shared_ptr<T> p(new Y);

...it is not necessary to have a virtual destructor when using shared_ptr. This is true whether T is void, or in the more familiar case where T is a polymorphic base of Y.

This goes against a long-standing conventional wisdom: That interface classes MUST have virtual destructors.

The OP's delete (void*) concern is solved by the fact that the shared_ptr constructor is a template that remembers the data type that it needs to destruct. This mechanism solves the virtual destructor problem in exactly the same way.

So even though the actual type of the object is not necessarily captured in the type of the shared_ptr itself (since T does not have to be the same type as Y), nevertheless, the shared_ptr remembers what type of object it is holding and it performs a cast to that type (or does something equivalent to that) when it comes time to delete the object.