我很难理解为什么不编译此代码。我将堆栈实现为双链表。我无法让我的AddToHead()工作。更具体地说,如果我尝试动态创建CharNode对象,则该程序将无法编译。我认为通过#include“ charlist.h”可以使程序访问CharNode类,因为它位于charlist.h中
我编译:g ++ -ansi -pedantic -Wall charlist.cxx -o clist
这是我得到的错误:
/tmp/ccHzaOmz.o: In function `CharList::AddToHead(char)':
charlist.cxx:(.text+0xe9): undefined reference to `CharNode::CharNode(char, CharNode*, CharNode*)'
collect2: error: ld returned 1 exit status
我知道未定义的引用意味着链接器找不到CharNode资源。我只是不知道为什么会在这里发生。
这是charlist.h
#ifndef __CharList__
#define __CharList__
#include <iostream>
#include <string>
using namespace std;
class CharList;
//CharNode class is clearly here in charlist.h
class CharNode
{
private:
char value;
CharNode* prev;
CharNode* next;
public:
CharNode(char value, CharNode* prev = NULL, CharNode* next = NULL);
friend class CharList;
};
class CharList
{
private:
CharNode* h;
CharNode* t;
public:
CharList();
~CharList();
bool IsEmpty() const;
char GetHead() const; //FUNCTION CAUSING ERROR
char GetTail() const;
void AddToHead(char v);
void AddToTail(char v);
};
#endif //__CharList__
这是charlist.cxx
#include <iostream>
#include <string>
#include <sstream>
#include <cassert>
#include <stdlib.h>
#include "charlist.h"
using namespace std;
CharList::CharList()
{
h = t = NULL;
}
bool CharList::IsEmpty() const
{
return (h == NULL);
}
//All other member functions excluded for relevancy
void CharList::AddToHead(char v){
CharNode* newHead;
newHead = new CharNode(v); //Why cant I do this? Error Line.
newHead->prev = NULL;
newHead->next = h;
if (IsEmpty()){
t = newHead;
h = newHead;
} else {
h->prev = newHead;
h = newHead;
}
}
最佳答案
在下面,您有一个构造函数的声明。这是一个承诺,您将在某处定义一个构造函数。
CharNode(char value, CharNode* prev = NULL, CharNode* next = NULL);
将其更改为也包含定义,您将不会得到该未定义的错误。
CharNode(char value, CharNode* prev = NULL, CharNode* next = NULL)
: value(value)
, prev(prev)
, next(next)
{
}