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

Question

I would like to have several types that share the same implementation but still are of different type in C++.

To illustrate my question with a simple example, I would like to have a class for Apples, Oranges and Bananas, all having the same operations and same implementation. I would like them to have different types because I want to avoid errors thanks to type-safety.

class Apple {
     int p;
public:
     Apple (int p) : p(p) {}
     int price () const {return p;}
}

class Banana {
     int p;
public:
     Banana (int p) : p(p) {}
     int price () const {return p;}
}

class Orange ...

In order not duplicating code, it looks like I could use a base class Fruit and inherit from it:

class Fruit {
     int p;
public:
     Fruit (int p) : p(p) {}
     int price () const {return p;}
}

class Apple: public Fruit {};
class Banana: public Fruit {};
class Orange: public Fruit {};

But then, the constructors are not inherited and I have to rewrite them.

Is there any mechanism (typedefs, templates, inheritance...) that would allow me to easily have the same class with different types?

Answer 1

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>;

Answer 2

Use templates, and use a trait per fruit, for example:

struct AppleTraits
{
  // define apple specific traits (say, static methods, types etc)
  static int colour = 0; 
};

struct OrangeTraits
{
  // define orange specific traits (say, static methods, types etc)
  static int colour = 1; 
};

// etc

Then have a single Fruit class which is typed on this trait eg.

template <typename FruitTrait>
struct Fruit
{
  // All fruit methods...
  // Here return the colour from the traits class..
  int colour() const
  { return FruitTrait::colour; }
};

// Now use a few typedefs
typedef Fruit<AppleTraits> Apple;
typedef Fruit<OrangeTraits> Orange;

May be slightly overkill! ;)

Answer 3

• C++11 would allow constructor inheritance: Using C++ base class constructors?
• Otherwise, you can use templates to achieve the same, e.g. template<class Derived> class Fruit;

Answer 4

There is also BOOST_STRONG_TYPEDEF.