Question

I read the Wikipedia article about the curiously recurring template pattern in C++ for doing static (read: compile-time) polymorphism. I wanted to generalize it so that I could change the return types of the functions based on the derived type. (This seems like it should be possible since the base type knows the derived type from the template parameter). Unfortunately, the following code won’t compile using MSVC 2010 (I don’t have easy access to gcc right now so I haven’t tried it yet). Anyone know why?

template <typename derived_t>
class base {
public:
    typedef typename derived_t::value_type value_type;
    value_type foo() {
        return static_cast<derived_t*>(this)->foo();
    }
};

template <typename T>
class derived : public base<derived<T> > {
public:
    typedef T value_type;
    value_type foo() {
        return T(); //return some T object (assumes T is default constructable)
    }
};

int main() {
    derived<int> a;
}

BTW, I have a work-around using extra template parameters, but I don’t like it—it will get very verbose when passing many types up the inheritance chain.

template <typename derived_t, typename value_type>
class base { ... };

template <typename T>
class derived : public base<derived<T>,T> { ... };

EDIT:

The error message that MSVC 2010 gives in this situation is error C2039: 'value_type' : is not a member of 'derived<T>'

g++ 4.1.2 (via codepad.org) says error: no type named 'value_type' in 'class derived<int>'

Answer 1

derived is incomplete when you use it as a template argument to base in its base classes list.

A common workaround is to use a traits class template. Here’s your example, traitsified. This shows how you can use both types and functions from the derived class through the traits.

// Declare a base_traits traits class template:
template <typename derived_t> 
struct base_traits;

// Define the base class that uses the traits:
template <typename derived_t> 
struct base { 
    typedef typename base_traits<derived_t>::value_type value_type;
    value_type base_foo() {
        return base_traits<derived_t>::call_foo(static_cast<derived_t*>(this));
    }
};

// Define the derived class; it can use the traits too:
template <typename T>
struct derived : base<derived<T> > { 
    typedef typename base_traits<derived>::value_type value_type;

    value_type derived_foo() { 
        return value_type(); 
    }
};

// Declare and define a base_traits specialization for derived:
template <typename T> 
struct base_traits<derived<T> > {
    typedef T value_type;

    static value_type call_foo(derived<T>* x) { 
        return x->derived_foo(); 
    }
};

You just need to specialize base_traits for any types that you use for the template argument derived_t of base and make sure that each specialization provides all of the members that base requires.