是否可以根据模板类型参数的嵌套typedef的存在来专门化模板定义？

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.

Answer 2

That's possible, and there are many ways to implement this. All of them should go back on some trait class has_type so that has_type<T>::value is true if the member typedef exists, and false otherwise. Let's assume we have this trait class already. Then here's one solution, using C++11 template aliases:

template <typename T, bool> class FooImpl
{
  // implement general case
};

template <typename T> class FooImpl<T, true>
{
  // implement specific case
};

template <typename T> using Foo = FooImpl<T, has_type<T>::value>;  // C++11 only

Now to make the typetrait:

template<typename T>
struct has_type
{
private:
    typedef char                      yes;
    typedef struct { char array[2]; } no;

    template<typename C> static yes test(typename C::context_type*);
    template<typename C> static no  test(...);
public:
    static const bool value = sizeof(test<T>(0)) == sizeof(yes);
};

If you don't have C++11, or if you don't want to rewrite the entire class, you can make the distinction more fine-grained, eg by using std::enable_if , std::conditional , etc. Post a comment if you want some specific examples.

Answer 3

Using @K-ballo's answer , I wrote the following:

namespace detail {
BOOST_MPL_HAS_XXX_TRAIT_DEF(context_type)
}

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

template <typename T>
class wrapper<T, typename boost::enable_if<detail::has_context_type<T> >::type>
{
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;
    }
};

Now, the sample code compiles and outputs:

T::context_type exists.
T::context_type does not exist.