从类到整数专门化模板

Question

I am playing with template specializations to learn their limits, and I was trying now not to specialize based on type , but using an integer parameter . But I am failing.

For instance, a template template <class T> should be specialized to have T for instance a string, but having an additional template parameter as template <int I> .

What does the standard say, and how can I do (if it can be done)? My code follows.

Thanks!

#include <iostream>
#include <typeinfo>
#include <tuple>
#include <string>

template <class T, class... U>
class many
{
public:

  T t;

  std::tuple<U...> u;
};


template <int size>
class many<int>
{
    // ???
};

int main(int argc, char* argv[])
{
  many<int, std::string, char> m;

  m.t = -1;

  std::get<0>(m.u) = "hello";
  std::get<1>(m.u) = 'w';

  std::cout << "many: " << std::endl;
  std::cout << m.t << std::endl;
  std::cout << std::get<0>(m.u) << std::endl;
  std::cout << std::get<1>(m.u) << std::endl;


  return 0;
}

Answer 1

Here's a way to specialize it for different integer values, it uses an extra type that you have to specialize further. It is pretty straightforward:

#include <iostream>
#include <memory>
#include <string>
#include <typeinfo>
#include <type_traits>

#include <string>

template <class T, class... U>
struct many
{
  T t;
  std::tuple<U...> u;
};

template<int N>
using Int = std::integral_constant<int, N>;

typedef Int <1> One;
typedef Int <2> Two;

template <>
struct many<int>
{ };

template <>
class many<One>
{ };

template <>
class many<Two>
{ };

int main(int argc, char* argv[])
{
  many<int, std::string, char> m;
  many<One, char> m2;
  m.t = -1;
  std::get<0>(m.u) = "hello";
  std::get<1>(m.u) = 'w';

  std::cout << "many: " << std::endl;
  std::cout << m.t << std::endl;
  std::cout << std::get<0>(m.u) << std::endl;
  std::cout << std::get<1>(m.u) << std::endl;
}