c++ - 基本模板继承成员，减少所需的输入

Question

我有一个社区已知的问题(可以找到最完整的答案之一 here )，但我想知道是否有办法让编译器理解你想要访问 base无需使用模板成员 typedef或 this如下:

#include <iostream>

template<typename T>
struct Base{
  T a;
  T a2;
};

template<typename T1, typename T2>
struct Derived : Base<T2>{
  T2 b;

  void member_fnc();
};

template<typename T1, typename T2>
void Derived<T1,T2>::member_fnc(){
  typedef Base<T2> base;
  std::cout << "a = " << base::a << std::endl;    // or this->a;
  std::cout << "a2 = " << base::a2 << std::endl;  // or this->a2;                     
  std::cout << "b = " << b << std::endl;
}

int main(){

  Derived<int,double> der1;
  der1.a = 1;
  der1.a2 = 2;
  der1.b = 1.1;
  der1.member_fnc();

}

我觉得应该有一种方法可以让编译器告诉您要访问基本模板 Base<T2>无需键入或包含 using对于您要访问的每个成员，因为您可以使用 using 取消嵌套命名空间关键字，但到目前为止我找不到。有可能吗？

Answer 1

为什么不呢:

template<typename T1, typename T2>
struct Derived : Base<T2>{
  typedef Base<T2> base;
  using base::a;
  using base::a2;

  T2 b;
  void member_fnc();
};

template<typename T1, typename T2>
void Derived<T1,T2>::member_fnc(){
  std::cout << "a = " << a << std::endl;    // or this->a;
  std::cout << "a2 = " << a2 << std::endl;  // or this->a2;                     
  std::cout << "b = " << b << std::endl;
}