How do I enable_if a class with variadic template arguments?

Question

Suppose I have a class with the following signature:

template <typename T, typename... Args>
class A;

But how this class behaves should depend on some other parameter, let's say it's the value of T::value:

template <typename T, typename... Args, typename Enable>
class A;

template <typename T, typename... Args, typename = typename std::enable_if<T::value>::type>
class A
{
  // do something
};

template <typename T, typename... Args, typename = typename std::enable_if<!T::value>::type>
class A
{
  // do something else
};

int main() { return 0; }

However, this program gives the following error:

prog.cpp:6:11: error: parameter pack ‘Args’ must be at the end of the template parameter list class A;

I have struggled to find a good source of information on the use of enable_if to select classes with variadic templates. The only question I could find is this one:

How to use std::enable_if with variadic template

But despite the name, this question and its answers aren't much help. If someone could provide or link a guide on how this should be approached and why that would be appreciated.

Answer 1

First of all, what you're trying is writing multiple definitions of a class template. That's not allowed because it violates One definition rule. If you want to do conditional enabling with classes, you need specializations. Also, the compiler error message already told you, you can't have a variadic parameter pack in the middle of a parameter list.

One way to do it would be:

namespace detail {

template<typename T, typename Enable, typename... Args>
class A_impl;

template<typename T, typename... Args>
class A_impl<T, typename std::enable_if<T::value>::type, Args...> {
    // code here
};

template<typename T, typename... Args>
class A_impl<T, typename std::enable_if<!T::value>::type, Args...> {
    // code here
};
}

template<typename T, typename...Args>
class A : public detail::A_impl<T, void, Args...> {};

Jonathan's way is also perfectly fine if the condition is really a bool, but it might not be useful if you wish to add more specializations that each depend on several conditons.

Answer 2

It looks as though for your purposes you don't need to enable/disable the class, you just need a partial specialization:

template <typename T, bool B = T::value, typename... Args>
  class A;

template <typename T, typename... Args>
  class A<T, true, Args...>;

template <typename T, typename... Args>
  class A<T, false, Args...>;