C ++-链表代码-在此范围内未声明LNode [英] C++ - Linked List code - LNode was not declared in this scope

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本文介绍了C ++-链表代码-在此范围内未声明LNode的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我正在用C ++编写链表的类,并且在写<<的运算符重载时遇到问题.我班的标题是:

I am writing a class for a linked list in C++ and I am having an issue when writing the operator overload for <<. My class' header is:

class LList
{
   private:
      struct LNode
      {
         LNode ();
         int data;
         LNode * next;
      };
   public:
      LList ();
      LList (const LList & other);
      ~LList ();
      LList & operator = (const LList & other);
      bool operator == (const LList & other);
      int Size () const;
      friend ostream & operator << (ostream & outs, const LList & L);
      bool InsertFirst (const int & value);
      bool InsertLast (const int & value);
      bool DeleteFirst ();
      bool DeleteLast ();
   private:
      LNode * first;
      LNode * last;
      int size;
};

而运算符是:

ostream & operator << (ostream & outs, const LList & L){ 
    LNode *np;
    np=L.first;
    while(np!=NULL){
        outs<<np->data;
        np=np->next;
    }
    return outs;
}

编译代码时出现错误:

LList.cpp: In function ‘std::ostream& operator<<(std::ostream&, const LList&)’:
LList.cpp:36:2: error: ‘LNode’ was not declared in this scope
  LNode *np;
  ^
LList.cpp:36:9: error: ‘np’ was not declared in this scope
  LNode *np;

我以为我可以在一个朋友函数中实例化一个结构,但这似乎不起作用.你们都知道发生了什么吗?

I thought I could instantiate a struct inside a friend function but it looks like that doesn't work. Any of you know what is happening?

推荐答案

由于您的operator<<是全局函数,因此需要使用以下命令访问内部类:

Since your operator<< is a global function, you need to access your inner class with:

LList::LNode *np;

由于此函数是LListfriend,因此它可以访问私有的LNode类.

Since this function is a friend of LList, then it can access the private LNode class.

这篇关于C ++-链表代码-在此范围内未声明LNode的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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