Is it possible to specialize a template on language linkage?

Question

A function's language linkage is part of its type:

7.5.1 [dcl.link] of the ISO C++ standard:

The default language linkage of all function types, function names, and variable names is C++ language linkage. Two function types with different language linkages are distinct types even if they are otherwise identical.

Is it possible to specialize a template on the type of a function pointer's linkage, or otherwise introspect the type of a function pointer to determine its linkage at compile time?

This first attempt does not seem legal:

#include <iostream>
#include <typeinfo>

struct cpp {};
struct c {};

extern "C++" void foo()
{
  std::cout << "foo" << std::endl;
}

extern "C" void bar()
{
  std::cout << "bar" << std::endl;
}

template<typename> struct linkage;

template<>
  struct linkage<void(*)()>
{
  typedef cpp type;
};

template<>
  struct linkage<extern "C" void(*)()>
{
  typedef c type;
}


int main()
{
  std::cout << "linkage of foo: " << typeid(linkage<decltype(&foo)>::type).name() << std::endl;
  std::cout << "linkage of bar: " << typeid(linkage<decltype(&bar)>::type).name() << std::endl;
  return 0;
}

g++-4.6 outputs:

$ g++ -std=c++0x test.cpp 
test.cpp:26:38: error: template argument 1 is invalid
test.cpp:26:3: error: new types may not be defined in a return type
test.cpp:26:3: note: (perhaps a semicolon is missing after the definition of ‘<type error>’)
test.cpp:32:10: error: two or more data types in declaration of ‘main’

Is there some application of SFINAE that could implement this functionality?

Answer 1

Yes, I believe that you should be able to specialize a template based on its language linkage according to the C++ standard. I tested the following code with the Comeau compiler online and it compiled with no errors:

#include <iostream>
#include <typeinfo>

struct cpp {};
struct c {};

extern "C++" typedef void(*cppfunc)();
extern "C" typedef void(*cfunc)();

extern "C++" void foo()
{
  std::cout << "foo" << std::endl;
}

extern "C" void bar()
{
  std::cout << "bar" << std::endl;
}

template<typename> struct linkage;

template<>
  struct linkage<cppfunc>
{
  typedef cpp type;
};

template<>
  struct linkage<cfunc>
{
  typedef c type;
};


int main()
{
  std::cout << "linkage of foo: " << typeid(linkage<decltype(&foo)>::type).name() << std::endl;
  std::cout << "linkage of bar: " << typeid(linkage<decltype(&bar)>::type).name() << std::endl;
  return 0;
}

However, I believe that due to a gcc bug, gcc does not distinguish function types based on language linkage, so this is not possible with gcc (and it doesn't seem sure when they will fix this).