c – 强制派生类型的注册

我缺少的是一种强制从Base派生的类通过初始化静态成员reg来注册自己的方法.换句话说,如果派生类没有定义/初始化静态成员,是否有可能让编译器以某种方式产生错误？

struct Registrar {
  Registrar(string type) {
    registry().push_back(type);
  }
  static vector<string> & registry() {
    static vector<string> * derivedTypes = new vector<string>;
    return *derivedTypes;
  }
};

//CRTP
template <typename Derived>
class Base
{
  static Registrar reg;
};

class Derived1 : public Base<Derived1> {/*Class definition*/};
class Derived2 : public Base<Derived2> {/*Class definition*/};
class Derived3 : public Base<Derived3> {/*Class definition*/};
//...

//Initialize the static members of each derived type
//Commenting out any of the following 3 lines doesn't produce an error.
//Ideally,I want it to produce a compile error.
template<> Registrar Base<Derived1>::reg("Derived1");
template<> Registrar Base<Derived2>::reg("Derived2");
template<> Registrar Base<Derived3>::reg("Derived3");

int main() {

  cout << "Registered Types:" << endl;
  for(vector<string>::const_iterator it = Registrar::registry().begin();
      it != Registrar::registry().end(); ++it) {
    cout << *it << endl;
  }

  return 0;
}