C++ covariant templates

Question

I feel like this one has been asked before, but I'm unable to find it on SO, nor can I find anything useful on Google. Maybe "covariant" isn't the word I'm looking for, but this concept is very similar to covariant return types on functions, so I think it's probably correct. Here's what I want to do and it gives me a compiler error:

class Base;
class Derived : public Base;

SmartPtr<Derived> d = new Derived;
SmartPtr<Base> b = d; // compiler error

Assume those classes are fully fleshed out... I think you get the idea. It can't convert a SmartPtr<Derived> into a SmartPtr<Base> for some unclear reason. I recall that this is normal in C++ and many other languages, though at the moment I can't remember why.

My root question is: what is the best way to perform this assignment operation? Currently, I'm pulling the pointer out of the SmartPtr, explicitly upcasting it to the base type, then wrapping it in a new SmartPtr of the appropriate type (note that this is not leaking resources because our home-grown SmartPtr class uses intrusive reference counting). That's long and messy, especially when I then need to wrap the SmartPtr in yet another object... any shortcuts?

Answer 1

Both the copy constructor and the assignment operator should be able to take a SmartPtr of a different type and attempt to copy the pointer from one to the other. If the types aren't compatible, the compiler will complain, and if they are compatible, you've solved your problem. Something like this:

template<class Type> class SmartPtr
{
    ....
    template<class OtherType> SmartPtr(const SmartPtr<OtherType> &blah) // same logic as the SmartPtr<Type> copy constructor

    template<class OtherType> SmartPtr<Type> &operator=(const SmartPtr<OtherType> &blah) // same logic as the SmartPtr<Type> assignment operator
};

Answer 2

SmartPtr<Base> and SmartPtr<Derived> are two distinct instantiations of a the SmartPtr template. These new classes do not share the inheritance that Base and Derived do. Hence, your problem.

what is the best way to perform this assignment operation?

 SmartPtr<Base> b = d; 

Does not invoke assignment operator. This invokes the copy-ctor (the copy is elided in most cases) and is exactly as if you wrote:

 SmartPtr<Base> b(d); 

Provide for a copy-ctor that takes a SmartPtr<OtherType> and implement it. Same goes for the assignment operator. You will have to write out the copy-ctor and op= keeping in mind the semantics of SmartPtr.

Answer 3

Templates are not covariant, and that's good; imagine what would happen in the following case:

vector<Apple*> va;
va.push_back(new Apple);

// Now, if templates were covariants, a vector<Apple*> could be
// cast to a vector<Fruit*>
vector<Fruit*> & vf = va;
vf.push_back(new Orange); // Bam, we just added an Orange among the Apples!

To achieve what you are trying to do, the SmartPointer class must have a templatized constructor, that takes either another SmartPointer or a pointer of another type. You could have a look at boost::shared_ptr, which does exactly that.

template <typename T>
class SmartPointer {

    T * ptr;

  public:
    SmartPointer(T * p) : ptr(p) {}
    SmartPointer(const SmartPointer & sp) : ptr(sp.ptr) {}

    template <typename U>
    SmartPointer(U * p) : ptr(p) {}

    template <typename U>
    SmartPointer(const SmartPointer<U> & sp) : ptr(sp.ptr) {}

    // Do the same for operator= (even though it's not used in your example)
};

Answer 4

Depends on the SmartPtr class. If it has a copy constructor (or in your case, assignment operator) that takes SmartPtr<T>, where T is the type it was constructed with, then it isn't going to work, because SmartPtr<T1> is unrelated to SmartPtr<T2> even if T1 and T2 are related by inheritance.

However, if SmartPtr has a templatized copy constructor/assignment operator, with template parameter TOther, that accepts SmartPtr<TOther>, then it should work.

Answer 5

Assuming you have control of the SmartPtr class, the solution is to provide a templated constructor:

template <class T>
class SmartPtr
{
    T *ptr;
public:

    // Note that this IS NOT a copy constructor, just another constructor that takes 
    // a similar looking class.
    template <class O>
    SmartPtr(const SmartPtr<O> &src)
    {
        ptr = src.GetPtr();
    }
    // And likewise with assignment operator.
};

If the T and O types are compatible, it will work, if they aren't you'll get a compile error.

Answer 6

I think the easiest thing is to provide automatic conversion to another SmartPtr according to the following:

template <class T>
class SmartPtr
{
public:
    SmartPtr(T *ptr) { t = ptr; }
    operator T * () const { return t; }
    template <class Q> operator SmartPtr<Q> () const
    { return SmartPtr<Q>(static_cast<Q *>(static_cast<T *>(* this))); }
private:
    T *t;
};

Note that this implementation is robust in the sense that the conversion operator template does not need to know about the semantics of the smart pointer, so reference counting does not need to replicated etc.