Templating a double-templated class method

Question

This is my simplified code :

template<typename device, template<typename device> class protocol>
class MyClass
{
public:
  template<typename select>
  bool method()
  {
    // Code
  }
};

I want method to act different in terms of the protocol type.

In other words, I have two different possible protocols, and I want to have two different behaviours for my method according to the protocols. But I don't know how to write it with templates.

Answer 1

By example, using SFINAE (if you accept a C++11 solution)

#include <iostream>
#include <type_traits>

template <typename>
struct protocol_1
 { };

template <typename>
struct protocol_2
 { };

template<typename device, template<typename> class protocol>
class MyClass
 {
   public:
      template<typename select, typename p = protocol<device>>
         typename std::enable_if<
         std::is_same<p, protocol_1<device>>::value, bool>::type
         method()
       { return true; }

      template<typename select, typename p = protocol<device>>
         typename std::enable_if<
         std::is_same<p, protocol_2<device>>::value, bool>::type
         method()
       { return false; }
 };

int main()
 {
   MyClass<int, protocol_1>  m1;
   MyClass<int, protocol_2>  m2;

   std::cout << m1.method<int>() << std::endl; // print 1 (true)
   std::cout << m2.method<void>() << std::endl; // print 0 (false)
 }

--- EDIT ---

As pointed by Yakk (thanks!), this solution is weak because use a template default value that can be explicited and circunvented.

An example; with

MyClass<int, protocol_1>{}.method<void>(); 

is called the "protocol_1" version of method(), using the default value for p; but explciting p, as follows

MyClass<int, protocol_1>{}.method<void, protocol_2<int>>(); 

is called the "protocol_2" version of method() over an istance of a MyClass based on protocol_1

To avoid this problem it's possible add a static_assert(), in both version of method(), to check and impose that p is equal to its default value (protocol<device>)

I mean... as follow

  template<typename select, typename p = protocol<device>>
     typename std::enable_if<
        std::is_same<p, protocol_1<device>>::value, bool>::type
     method()
   {
     static_assert(std::is_same<p, protocol<device>>::value, "!");

     return true;
   }

  template<typename select, typename p = protocol<device>>
     typename std::enable_if<
     std::is_same<p, protocol_2<device>>::value, bool>::type
     method()
   {
     static_assert(std::is_same<p, protocol<device>>::value, "!");

     return false;
   }

So

MyClass<int, protocol_1>{}.method<void, protocol_2<int>>();

generate a compiler error.