How to define different types for the same class in C++
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?
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>;
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! ;)
- 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;
There is also BOOST_STRONG_TYPEDEF.