Is it possible to specialize a template definition based on the existence of a nested typedef of a template type parameter?

Question

I have a template, template <typename T> class wrapper, that I would like to specialize based on the existence of typename T::context_type. If typename T::context_type is declared, then the constructors and assignment operator overloads of the wrapper<T> instantiation should accept a mandatory typename T::context_type parameter. Additionally, wrapper<T> objects would store "context" in the member data. If typename T::context_type does not exist, then the constructors and assignment operator overloads of wrapper<T> would take one less parameter and there would be no additional data member.

Is this possible? Can I get the following code to compile without changing the definitions of config1, config2, and main()?

#include <iostream>

template <typename T, bool context_type_defined = true>
class wrapper
{
public:
    typedef typename T::context_type context_type;

private:
    context_type ctx;

public:
    wrapper(context_type ctx_)
        : ctx(ctx_)
    {
        std::cout << "T::context_type exists." << std::endl;
    }
};

template <typename T>
class wrapper<T, false>
{
public:
    wrapper() {
        std::cout << "T::context_type does not exist." << std::endl;
    }
};

class config1 {
public:
    typedef int context_type;
};

class config2 {
public:
};

int main()
{
    wrapper<config1> w1(0);
    wrapper<config2> w2;
}

Answer 1

Yes, it is possible. I have implemented such behavior in the past by using some metaprogramming tricks. The basic build blocks are:

BOOST_MPL_HAS_XXX_TRAIT_DEF, to define a metafunction predicate that will evaluate to a true type if the argument is of class type and has a nested type with a given name (context_type in your case).

http://www.boost.org/doc/libs/1_47_0/libs/mpl/doc/refmanual/has-xxx-trait-def.html

Boost.EnableIf, to define the specializations based on the previously defined trait.

http://www.boost.org/libs/utility/enable_if.html # See 3.1 Enabling template class specializations

---

Note that you may be able to get that behavior working directly with SFINAE, something like this may work:

template< typename T, typename Context = void >
class wrapper { ... }; // Base definition

template< typename T >
class wrapper< T, typename voif_mfn< typename T::context_type >::type > { ... }; // Specialization

However, I like the expressiveness of the solution based on traits and enable if.