C++ Template Specialization: compile error: "is not a type"

Question

If I remove the template specialization part (the one that tries to print "Test 2"), the code compiles fine, but I'd like to be able to have a special case that runs a different code path that looks clean to outside users.

#include <iostream>

using namespace std;

struct SpecialType {};

template<typename A , typename B = SpecialType>
class Test
{
public:
    class TestInner
    {
    public:
        TestInner& operator* ();
    };
};

template<typename A , typename B>
typename Test<A , B>::TestInner& Test<A , B>::TestInner::operator* ()
{
    cout << "Test 1" << endl;
    return *this;
}

// If the following is removed, everything compiles/works, but I want this alternate code path:
template<typename A>
typename Test<A , SpecialType>::TestInner& Test<A , SpecialType>::TestInner::operator* ()
{
    cout << "Test 2" << endl;
    return *this;
}

int main()
{
    Test<int , SpecialType>::TestInner test;
    *test;

   return 0;
}

What am I doing wrong?

Edit: By the way, the compiler error reads:

main.cpp:26:44: error: 'Test<A, SpecialType>::TestInner' is not a type
 typename Test<A , SpecialType>::TestInner& Test<A , SpecialType>::TestInner::operator* ()
                                            ^
main.cpp:26:89: error: invalid use of dependent type 'typename Test<A, SpecialType>::TestInner'
 typename Test<A , SpecialType>::TestInner& Test<A , SpecialType>::TestInner::operator* ()
                                                                                         ^

Answer 1

Add a declaration for the specialised class:

template<typename A>
class Test<A, SpecialType>
{
public:
    class TestInner
    {
    public:
        TestInner& operator* ();
    };
};

The problem is that you define an member for a specialization that is not declared. A specialization of a templated class does not share any members or methods with the generalised template, so the declaration of the generalised template doesn't serve as a declaration of any specialisations of that template class.

Consider this:

template <class T>
class Foo {
  void GeneralFunction(T x);
}

and a specialization:

template <>
class Foo<int> {
  void SpecialisedFunction(int x);
}

Here, Foo</*anything except int*/> has only the method GeneralFunction whereas Foo<int> has only the method SpecialisedFunction.

By the same logic, this is allowed also:

template<>
class Foo<float> {
  float GeneralFunction; //here GeneralFunction is a data member, not a method.
}

Long story short you need to declare you specialisation.