C++ template C2039 - c++

Question

This compiles:
template <class T>
class Bar {};
template<class T, class Dummy=void>
class Foo;
template<class T>
class Foo <T, typename std::enable_if<
std::is_base_of<Bar<T>, T>::value
>::type> {
public:
//THIS CHANGES IN THE 2ND SNIPPET
void test () const {
std::cout << "test";
}
};
class Cat : Bar<Cat> {};
int main () {
Foo<Cat> foo;
foo.test();
return 0;
}
This errors:
template <class T>
class Bar {};
template<class T, class Dummy=void>
class Foo;
template<class T>
class Foo <T, typename std::enable_if<
std::is_base_of<Bar<T>, T>::value
>::type> {
public:
//THIS CHANGED!
void test () const;
};
//THIS WAS ADDED SINCE THE 1ST SNIPPET!
template<class T>
void Foo<T>::test () const {
std::cout << "test";
} //error C2039: 'test' : is not a member of 'Foo<T>'
class Cat : Bar<Cat> {};
int main () {
Foo<Cat> foo;
foo.test();
return 0;
}
I have marked the differences. Why does it error in the 2nd code snippet? How do I keep declaration and definition separate while avoiding the error?
I'm guessing it's got something to do with this:
"template
void Foo::test () const"
Like, this is the wrong way to tell the compiler that the method test() const is a method of template class template class Foo , T>::value >::type>
I've, of course, looked it up on Google and StackOverflow but it seems that, whenever this error pops up for templates, it's a different reason every time. (Probably because a lot of things can cause the C2039 error.)
Also, could a mod. or someone help me add the C2039 tag to this post? It says I need a min. of 1500 rep. to add that tag.
-- Rambling --
It also be noted that it's been a while since I've used C++; and even longer since I've used templates. I know this might be a weird way to use templates but I can assure you I have a valid reason!

Answer 1

An example of out-of-class member declaration for a partially specialized class template is given in 14.5.4.3/1 (C++03). And this is what it looks like
// primary template
template<class T, int I> struct A {
void f();
};
// class template partial specialization
template<class T> struct A<T,2> {
void g();
};
// member of class template partial specialization
template<class T> void A<T,2>::g() { }
As you can see, you have to specify specialized arguments in the out-of-class member definition.
In your case it should be
template<class T>
void Foo<T, typename std::enable_if<
std::is_base_of<Bar<T>, T>::value>::type>::test () const {
std::cout << "test";
}