c++ - 嵌套从属名称在成员初始化列表如何被分别解析为类型或数据成员？

Question

我在看 Effective C++ 第四十二条，它指出 C++ 在解析嵌套从属名称（nested dependent name）时并不会优先假定它是「类型」，于是得显式地在所有嵌套从属类型名称前统统加上 typename, 以声明它是嵌套从属类型名称（nested dependent type name）。但是这 typename 不能用在成员初始化列表（member initialization list）里。问题来了：

假定 varialble_ 是 Base<T> 的数据成员，类型为 int. 那么

template<typename T>
class Derived: public Base<T> {
 public:
  explicit Derived(int x) : Base<T>::variable_(x) {}
}

按照 C++ 的解析规则，Base<T>::variable_ 应该会被优先解析成非类型的名称，即 Base<T> 的一个数据成员，从而在 Derived 构造函数里被初始化成 x.

但是这么一来，假定 Base<T>::Nested是一种 class, 则

template<typename T>
class Derived: public Base<T> {
 public:
  explicit Derived(int x) : Base<T>::Nested(x) {}

 private:
  int second_variable_;
}

怎么还能保证 Base<T>::Nested(x) 会调用 Base<T>::Nested(int x) 构造函数来初始化其类对象，而不是把 Base<T>::Nested 当成一种 Base<T> 的数据成员呢？

Answer 1

I suddenly realized that I had confused the constructors of Base<T>::Nested(x) and Base<T>. Derived As a derived class, it can only call the constructor of the base class Base<T> and initialize its own data members within the member initialization list.

If Derived inherits Base<T>::Nested, this actually means that Base<T>::Nested is a class , and the compiler will know that Base<T>::Nested(int x) in the member initialization list is a constructor. In fact, Effective C++ Article 42 also points out that C++ prohibits the use of typename in base class lists. After all, it is superfluous.

Having said that, in the member initialization list, Base<T>::variable_ needs to be parsed into a type, does not make sense.