How to define different types for the same class in C++

A common technique is to have a class template where the template argument simply serves as a unique token (“tag”) to make it a unique type:

template <typename Tag>
class Fruit {
    int p;
public:
    Fruit(int p) : p(p) { }
    int price() const { return p; }
};

using Apple = Fruit<struct AppleTag>;
using Banana = Fruit<struct BananaTag>;

Note that the tag classes don’t even need to be defined, it’s enough to declare a unique type name. This works because the tag isn’s actually used anywhere in the template. And you can declare the type name inside the template argument list (hat tip to @Xeo).

The using syntax is C++11. If you’re stuck with C++03, write this instead:

typedef Fruit<struct AppleTag> Apple;

---

If the common functionality takes up a lot of code this unfortunately introduces quite a lot of duplicate code in the final executable. This can be prevented by having a common base class implementing the functionality, and then having a specialisation (that you actually instantiate) that derives from it.

Unfortunately, that requires you to re-implement all non-inheritable members (constructors, assignment …) which adds a small overhead itself – so this only makes sense for large classes. Here it is applied to the above example:

// Actual `Fruit` class remains unchanged, except for template declaration
template <typename Tag, typename = Tag>
class Fruit { /* unchanged */ };

template <typename T>
class Fruit<T, T> : public Fruit<T, void> {
public:
    // Should work but doesn’t on my compiler:
    //using Fruit<T, void>::Fruit;
    Fruit(int p) : Fruit<T, void>(p) { }
};

using Apple = Fruit<struct AppleTag>;
using Banana = Fruit<struct BananaTag>;