如何将类型添加为类型特征

Question

I'd like to have a type trait where I associate a class that implements an interface class with the interface class. For example, consider there is an abstract base class and a concrete implementation

class IFoo
{
public:
    virtual ~IFoo() = default;
    virtual void doStuff() = 0;
    // more abstract base class stuff
};

class ConcreteFoo: public IFoo
{
public:
    void doStuff() override;
    // concrete class stuff
};

Now, I am looking for a way to use a type trait to get a type that's a concrete implementation of IFoo so that the following would be possible:

using IFooImpl = get_implementation_of<IFoo>::type;
std::unique_ptr<IFoo> foo = std::make_unique<IFooImpl>();

Does anyone know if this is possible with C++ type traits?

Answer 1

There is no standard way to do it. If there're many subclasses extends from the special class, which one should be returned?

class task {
public:
  
  virtual void execute() = 0;
};

class hello_world_task : 
  public virtual task {
public:
  
  virtual void execute() override {

    std::cout << "Hello World!" << std::endl;
  }
};

class exit_task : 
  public virtual task {
public:

  virtual void execute() override {

    std::exit(0);
  }
};

auto fun() {

  // `exit_task` or `hello_world_task` ?
  using task_impl = get_implementation_of<task>::type;
}

It's necessary to register implementation class by some means. There's a simple way:

template <typename T>
struct get_implementation_of {
};

template <>
struct get_implementation_of<task> {
  using type = hello_world_task;
};

auto fun() {

  // task is hello_world_task
  using task_impl = get_implementation_of<task>::type;
  task_impl().execute();
}

so you can:

// register ConcreteFoo as the implementation class of IFoo
template <>
struct get_implementation_of<IFoo> {
  using type = ConcreteFoo;
};

auto fun() {

  using IFooImpl = get_implementation_of<IFoo>::type;
  std::unique_ptr<IFoo> foo = std::make_unique<IFooImpl>();
}