如果对象是堆栈创建(包括继承类型),是否可以发出编译错误？

Question

基本上我希望通过工厂方法创建所有子类型(我有一个高大的域层次结构,有200多个类).

因为new,这不是问题,因为这可以在A(new私有)中被覆盖.

class A{
protected:
  A();
public:
  template<class T, typename... ARGUMENTS>
  static T* create(ARGUMENTS&&... arguments);
};

class B : public A {
public:
  B();
};

void test() {
  B b;//compile error wanted here - but as a consequence of inheriting A
}

这里A是"库/框架"类.而B是"用户创建的类".在B上要求typedef或类似可能没问题.

更新:我在A上添加了'create'函数,我打算用它来创建对象.

Answer 1

你可以在构造中要求一个令牌,A只在身体中传递A::create

#include <utility>

class A{
private:
  struct create_token
  { 
    create_token(const create_token &) = delete; 
    create_token& operator=(const create_token &) = delete; 
    create_token(create_token &&) = default; 
    create_token& operator=(create_token &&) = default; 
  };
protected:
  A(create_token) {}
public:
  template<class T, typename... ARGUMENTS>
  static T* create(ARGUMENTS&&... arguments)
  {
    // Whatever creation mechanism here
    return new T(create_token{}, std::forward<ARGUMENTS>(arguments)...);
  }
};

class B : public A {
public:
  template <typename Token> // Can't name A::create_token, it is private
  B(Token tok) : A(std::move(tok)) {}
  B(){} // Will always lack a `create_token`
};

int main() {
  B b;//compile error wanted here - but as a consequence of inheriting A
  B* b = A::create<B>();
}

现场观看