C ++：实现在匿名结构中定义的函数 [英] C++: Implementing a function defined inside an anonymous structure

问题描述

给定以下代码：

class Named {
  class /*Unnamed*/ {
    void Function();
  } un;
};

// Implement Named::Unnamed::Function here

int main() {
  Named named;
  named.un.Function();
}

有没有任何方法来实现Named :: Unnamed :: Function未命名或嵌入函数的定义在Named的定义内？

Is there any way to implement Named::Unnamed::Function without either naming Unnamed or embedding the function's definition within the definition of Named?

我猜测答案是no，但GCC给了我有用的消息 Named :: {unnamed type＃2} :: Function（）'，它发生在我可能有一些疯狂的语法。

I'm guessing the answer is "no", but GCC gives me the useful message "undefined reference to `Named::{unnamed type#2}::Function()', and it occured to me there might be some crazy possible syntax.

推荐答案

这实际上是可能的（在C ++ 11中），有两种方式：

This is actually possible (in C++11), and in two ways:

struct Named {
  struct /*Unnamed*/ {
    void Function();
  } un;
  typedef decltype(un) Unnamed;
};

// #1
void Named::Unnamed::Function(){
}

//// #2
//typedef decltype(Named::un) Unnamed;
//void Unnamed::Function(){
//}
//// same way, using template alias
//template<class T> using alias = T;
//void alias<decltype(Unnamed::un)>::Function(){
//}

int main() {
  Named named;
  named.un.Function();
}

这可以感谢 $ 9.1 [class.name] p5 ：

定义类类型或cv限定版本的 typedef-name （7.1.3）

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