C++ static polymorphism (CRTP) and using typedefs from derived classes

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.

One small drawback of using traits is that you have to declare one for each derived class. You can write a less verbose and redondant workaround like this :

template <template <typename> class Derived, typename T>
class base {
public:
    typedef T 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;
}

In C++14 you could remove the typedef and use function auto return type deduction:

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

This works because the deduction of the return type of base::foo is delayed until derived_t is complete.

An alternative to type traits that requires less boilerplate is to nest your derived class inside a wrapper class that holds your typedefs (or using's) and pass the wrapper as a template argument to your base class.

template <typename Outer>
struct base {
    using derived = typename Outer::derived;
    using value_type = typename Outer::value_type;
    value_type base_func(int x) {
        return static_cast<derived *>(this)->derived_func(x); 
    }
};

// outer holds our typedefs, derived does the rest
template <typename T>
struct outer {
    using value_type = T;
    struct derived : public base<outer> { // outer is now complete
        value_type derived_func(int x) { return 5 * x; }
    };
};

// If you want you can give it a better name
template <typename T>
using NicerName = typename outer<T>::derived;

int main() {
    NicerName<long long> obj;
    return obj.base_func(5);
}