將文件讀入鏈接列表在一個類中使用節點結構的C++

Question

我試圖將包含字母「farming」的文本文件讀入鏈接的節點列表。我創建了一個名爲NumberList的類，它具有節點的結構。這是標題。

#ifndef NUMBERLIST 
#define NUMBERLIST 
#include <iostream> 

using namespace std; 

class NumberList 
{ 
protected: 
//declare a class for the list node 
//constructor to initialize nodes of list 
struct ListNode 
{ 
    char value; 
    ListNode *next; 

    // Constructor 
    ListNode(char value1, ListNode *next1 = NULL) 
    { 
     value = value1; 
     next = next1; 
    } 
}; 

ListNode *head; //pointer to head of the list 

public: 
NumberList() { head = NULL; } //constructor 
~NumberList();  //destructor 
void displayList() const; //print out list 
void reverse(); 

}; 
#endif 

我遇到問題的地方是嘗試將文本文件讀入main（）中的鏈接列表中。

下面是我在主：

#include "Numberlist.h" 
#include "ReliableNumberList.h" 
#include <iostream> 
#include <fstream> 


using namespace std; 

int main() 
{ 

ListNode *letterList = nullptr; //create a linked list 
char letter; 
        //This is where I read the file into the list 
//open the file 
ifstream letterFile("linkedText.txt"); 
if (!letterFile) 
{ 
    cout << "Error in opening the file of letters."; 
    exit(1); 
} 
//read the file into a linked list 
while (letterFile >> letter) 
{ 
    //create a node to hold this letter 
    letterList = new ListNode(letter, letterList); 
    //missing a move to the next node? 
} 
return 0; 
} 

該讀文件樣本從我的課本來了，但結構讀碟到沒有位於一個單獨的類。對於我的生活，我無法弄清楚如何在NumberList類中引用ListNode結構體。 Visual Studio指出ListNode和letterList是未定義的。我知道它是因爲我沒有從NumberList類中正確地引用它們。

任何幫助將不勝感激。

Answer 1

迅速解決你的問題可能是這樣的：

//------------------------------NumberList.hpp----------------------------- 

#ifndef NUMBERLIST_HPP 
#define NUMBERLIST_HPP 
#include <iostream> 

class NumberList{ 
protected: 
    //Protected Members can't be used outside the class 
    struct ListNode{ 
     char value; 
     ListNode *next; 
     ListNode(char value1, ListNode *next1 = NULL){ 
      value = value1; 
      next = next1; 
     } 
    }; 
    ListNode *head, *tail; //class members 
    //head always points at the 1st letter, tail is used for quick adding at the end 
public: 
    NumberList() { head = NULL; tail = NULL; } 
    ~NumberList(); //don't forget to deallocate space properly at the end 
    void displayList() const; //print out list 
    void reverse(); 
    void add(char newchar) { 
     //allocate a new node using the ListNode constructor, by default, next1 will be null 
     ListNode *newNode = new ListNode(newchar); //equvalent to ListNode(newchar, NULL); 
     if (tail == NULL) { //if no elements in the list, both show to newNode 
      tail = newNode; 
      head = newNode; 
     }else{ 
      tail->next = newNode; //make last node -> next pointer, point to newNode (new last node) 
      tail = tail->next; //make current last node be the actual last node 
     } 
    } 
}; 
#endif 

//------------------------------Main.cpp----------------------------- 
#include "Numberlist.hpp" 
#include <iostream> 
#include <fstream> 

using namespace std; 

int main(){ 
    ifstream letterFile("linkedText.txt"); 
    if (!letterFile){ 
     cout << "Error in opening the file of letters."; 
     exit(-1); 
    } 

    NumberList numberList; 

    char letter; 
    while (letterFile >> letter) numberList.add(letter); 
} 

稍微改變你的邏輯，你不再添加列表節點列表， 但我懷疑你不想，無論如何。相反，最好將字符 直接添加到列表中，並讓列表處理它的節點（因爲節點結構被保護，所以是合理的）。

當然，類需要更精煉，但是這應該解決您最初的問題。