Passing a derived class to a template function specialized with base class

Question

The following code compiles fine, but produces a linker error:

class Base {};

class Derived : public Base {};

template 
void f(const T&am         


        

Answer 1

The linker error occurs because you haven't defined, only declared, the function template

template 
void f(const T& value);

To avoid it, define the above template and your code will compile, but presumably it still doesn't do what you want.

When calling f with an argument of type Derived the above template is a better match compared to your specialization because the latter requires a derived-to-base conversion while the former doesn't.

You can achieve the behavior you want by using enable_if to allow the first template to participate in overload resolution only if the deduced template argument type is not Base or a type derived from Base.

template 
typename std::enable_if::value>::type
    f(const T& value) {

}

And change the other f so it's not a specialization, but just an overload.

void f(const Base& value) {
    // ...
}

Live demo

In general, prefer overloading function templates over specialization. Read this to learn about the pitfalls of function template specialization.