CODEWITHSUNDEEP
c++static
Kazade
Kazade

Reputation: 1327

C++ compile-time and runtime available name attribute

I have a Controller class which is designed to be inherited. It currently has a name() pure virtual method, like this:

class Controller {
public:
    virtual const std::string name() const = 0;
}

The name is really a class-property, rather than an instance property. It shouldn't change between instances of a subclass, so really, virtual isn't the best thing to do here. Anyway, I've now hit a problem where I need the name at compile-time, in a template function. Obviously T::name() doesn't work, and because the subclasses won't be default constructible, I can't even hack around it with T()::name().

So something needs to change, but I can't figure out the best way to do this. I need to be able to access name in the following situations:

name() doesn't have to be method, it can be a public attribute, or something else. But no-matter what I try I can only seem to satisfy 2 of the above 3 requirements.

Can anyone suggest any good approaches for solving this?

Upvotes: 0

Views: 244

Answers (1)

Serge Kireev
Serge Kireev

Reputation: 176

Unfortunately, you cannot declare a static method virtual so something like this wouldn't work :

struct BaseClass {
  static virtual std::string name() =0;
};

struct ChildClass  {
  static std::string name() {
    return "my_name";
  }
};

you can instead make a method with a slightly different name :

struct BaseClass {
  virtual std::string name() =0;
};

struct ChildClass  {
  std::string name() {
    return ChildClass::_name(); 
  };

  static std::string _name() {
    return "my_name";
  }
};

which will work with the three cases you describe :

ChildClass c;
BaseClass& b = c;
std::cout << ChildClass::_name() << std::endl;
std::cout << c.name() << std::endl;
std::cout << b.name() << std::endl;

Upvotes: 1

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