c++ - 为什么在作为值返回时在类内部定义结构需要范围解析？

Question

我有一个单向链表实现，如下所示:

标题

class SinglyLinkedList
{
  struct Node
  {
    Node * _pNext;
    int    _data;
  };

public:

  Node * SomeFun(Node * ip1, Node * ip2);  
  // Some more methods here
};

现在在实现这个类的其中一个方法的时候

CPP

Node * SinglyLinkedList::SomeFun(Node * ip1, Node * ip2)
{
  //Some code and return
}

我不理解的奇怪行为是编译时，编译器 拒绝识别返回类型中的类型“节点”，除非我将其指定为 SinglyLinkedList::Node。但是函数参数的相同类型被识别 没有明确指定..理想情况下，我觉得在这两种情况下都应该有 不需要这个显式指定，因为 Node 是在同一个类中定义的。 任何人都可以对此有所了解吗？

Answer 1

SinglyLinkedList::Node * SinglyLinkedList::SomeFun

在这里，您不在类范围内。但是在参数子句中，或者在函数中，你在类范围内，所以你不应该限定 Node 来自 SinglyLinkedList 类，因为编译器已经知道了。

n3376 3.3.7/1

The following rules describe the scope of names declared in classes.

The potential scope of a declaration that extends to or past the end of a class definition also ex- tends to the regions defined by its member definitions, even if the members are defined lexically outside the class (this includes static data member definitions, nested class definitions, member func- tion definitions (including the member function body and any portion of the declarator part of such definitions which follows the declarator-id, including a parameter-declaration-clause and any default arguments