[英]C++ Node class, linked list, LIFO, FIFO
我必须写一个“程序员友好”的链表。 我有两个类,您的主类和节点类。 我的节点类将是我的数据结构,并且必须包含下面列出的信息。 所有节点信息都将保存在节点类中。 我的主类应该只是一个调用节点函数的驱动程序。
主类应包含:
您的 Node 类应该执行以下操作:
笔记:
我的 Node.cpp 代码如下所示:它抛出此错误:抛出异常:读取访问冲突。 这是 nullptr。
#include "stdafx.h"
#include "Node.h"
#include "iomanip"
Node::Node()
{
}
Node::~Node()
{
}
void Node::AppendFIFO(int id, string userName, string password, string fName, string lName)
{
}
void Node::AppendLIFO(int id, string userName, string password, string fName, string lName)
{
//cout << "Yo" << endl;
Node *curNode;
curNode = head;
cout << "user name:" << userName.length();
cout << "Password:" << password.length();
cout << "f name:" << fName.length();
cout << "l name:" << lName.length();
while (curNode != nullptr)
{
curNode = curNode->next;
}
}
void Node:: display()
{
}
我的 Node.h 文件看起来像这样
#pragma once
#include "iostream"
#include "string"
#include "iomanip"
using namespace std;
class Node
{
public:
Node();
~Node();
void AppendFIFO(int, string, string, string, string);
void AppendLIFO(int, string, string, string, string);
void display();
private:
int ID;
string userName;
string password;
string firstName;
string lastName;
Node *next;
Node *head;
};
我的 main.cpp 文件代码如下所示
#include "stdafx.h"
#include "iostream"
#include "string"
#include "Node.h"
#include "iomanip"
using namespace std;
int main()
{
Node *LIFO = nullptr;
Node *FIFO = nullptr;
//The 10 users information for LIFO
LIFO->AppendLIFO(10, "postMalone", "Asdkc34D", "Austin", "Post");
LIFO->AppendLIFO(9, "h20", "akdjW78v", "Benny", "Washington");
LIFO->AppendLIFO(8, "testing", "aklc5kaS", "Timmy", "Trump");
LIFO->AppendLIFO(7, "Rob-by", "robby3939", "Robert", "Malone");
LIFO->AppendLIFO(6, "TracyLMoore", "Moore098", "Tracy", "Moore");
LIFO->AppendLIFO(5, "Billybill", "Bb234", "Bill", "Prescott");
LIFO->AppendLIFO(4, "beth-09", "09ASDN", "Beth", "Richards");
LIFO->AppendLIFO(3, "Gabe123", "123ilkkSW", "Gabriel", "Smith");
LIFO->AppendLIFO(2, "sthomas", "sthom56712", "Shannon", "Thomas");
LIFO->AppendLIFO(1, "bob", "LKJG840", "Bobby", "Steve");
LIFO->AppendLIFO(0, "zmoore00", "00SDJ", "Zackary", "Moore");
//The 10 users information for FIFO
FIFO->AppendFIFO(10, "postMalone", "Asdkc34D", "Austin", "Post");
FIFO->AppendFIFO(9, "h20", "akdjW78v", "Benny", "Washington");
FIFO->AppendFIFO(8, "testing", "aklc5kaS", "Timmy", "Trump");
FIFO->AppendFIFO(7, "Rob-by", "robby3939", "Robert", "Malone");
FIFO->AppendFIFO(6, "TracyLMoore", "Moore098", "Tracy", "Moore");
FIFO->AppendFIFO(5, "Billybill", "Bb234", "Bill", "Prescott");
FIFO->AppendFIFO(4, "beth-09", "09ASDN", "Beth", "Richards");
FIFO->AppendFIFO(3, "Gabe123", "123ilkkSW", "Gabriel", "Smith");
FIFO->AppendFIFO(2, "sthomas", "sthom56712", "Shannon", "Thomas");
FIFO->AppendFIFO(1, "bob", "LKJG840", "Bobby", "Steve");
FIFO->AppendFIFO(0, "zmoore00", "00SDJ", "Zackary", "Moore");
system("pause");
return 0;
}
我无法弄清楚错误有什么问题。 我也无法弄清楚如何确定字符串组中最长字符串的长度,然后将其存储以便能够用于 setw 以进行正确显示。请帮助。
它抛出此错误:抛出异常:读取访问冲突。 这是 nullptr。
你的程序开始于
int main() { Node *LIFO = nullptr; ... LIFO->AppendLIFO(10, "postMalone", "Asdkc34D", "Austin", "Post");
在Node::AppendLIFO你做
curNode = head;
所以你尝试在这是nullptr 时访问head ,然后你使用未定义的结果来做
curNode = curNode->next;
目前您的程序从未创建Node的实例,它不能像那样工作
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