Visibility of template specialization of C++ function

Question

Suppose I have fileA.h which declares a class classA with template function SomeFunc(). This function is implemented directly

Answer 1

I had the same problem with gcc4, here is how i solved it. It was more simple a solution than what i was lead to believe by previous comments. The previous posts ideas were correct but their syntax didn't work for me.

    ----------header-----------------
    template < class A >
    void foobar(A& object)
    {
      std::cout << object;
    }

    template <> 
    void foobar(int);

    ---------source------------------
    #include "header.hpp"

    template <>
    void foobar(int x)
    {
      std::cout << "an int";
    }