Compiler warning at C++ template base class

Question

I get a compiler warning, that I don't understand in that context. When I compile the "Child.cpp" from the following code. (Don't wonder: I stripped off my class declarations to the bare minimum, so the content will not make much sense, but you will see the problem quicker). I get the warning with Visual Studio 2003 and Visual Studio 2008 on the highest warning level.

---

The code

AbstractClass.h:

#include <iostream>

template<typename T>
class AbstractClass
{
    public:
        virtual void Cancel(); // { std::cout << "Abstract Cancel" << std::endl; };
        virtual void Process() = 0;
};

// Outside definition. If I comment out this and take the inline
// definition like above (currently commented out), I don't get
// a compiler warning.
template<typename T>
void AbstractClass<T>::Cancel()
{
    std::cout << "Abstract Cancel" << std::endl;
}

Child.h:

#include "AbstractClass.h"

class Child : public AbstractClass<int>
{
    public:
        virtual void Process();
};

Child.cpp:

#include "Child.h"
#include <iostream>

void Child::Process()
{
    std::cout << "Process" << std::endl;
}

---

The warning

The class "Child" is derived from "AbstractClass". In "AbstractClass" there's the public method "AbstractClass::Cancel()". If I define the method outside of the class body (like in the code you see), I get the compiler warning...

AbstractClass.h(7) : warning C4505: 'AbstractClass::Cancel' : unreferenced local function has been removed with [T=int]

...when I compile "Child.cpp". I do not understand this, because this is a public function, and the compiler can't know if I later reference this method or not. And, in the end, I reference this method, because I call it in main.cpp and despite this compiler warning, this method works if I compile and link all files and execute the program:

//main.cpp
#include <iostream>
#include "Child.h"

int main()
{
    Child child;
    child.Cancel();  // Works, despite the warning
}

If I do define the Cancel() function as inline (you see it as out commented code in AbstractClass.h), then I don't get the compiler warning. Of course my program works, but I want to understand this warning or is this just a compiler mistake?

Furthermore, if do not implement AbsctractClass as a template class (just for a test purpose in this case) I also don't get the compiler warning...?

---

If I make a non-virtual function, I don't get the compile warning for that non-virtual function, but all answers up to now don't comprise the virtual stuff. Try this:

template<typename T>
class AbstractClass
{
    public:
        virtual void Cancel(); // { std::cout << "Abstract Cancel" << std::endl; };
        virtual void Process() = 0;
        void NonVirtualFunction();
};

//...

template<typename T>
void AbstractClass<T>::NonVirtualFunction()
{
    std::cout << "NonVirtualFunction" << std::endl;
}

The answers up to know helped me, but I don't think that the question is fully answered.

Answer 1

The warning simply says that the linker cannot see any usage of the function.

If you want to"tell" the linker to avoid the warning you can fool the linker into "think" it is in use.

For example:

void MyLinkerThinkNotUsedFunction
{
}

void* Foo = (void*)MyLinkerThinkNotUsedFunction;

will suffice to avoid any warning C4505 on the function MyLinkerThinkNotUsedFunction.