c++ - 链表更改节点而不是添加另一个节点

Question

理论上，下面的程序应该创建一个链表，然后用户可以添加、显示或减去(不相关)。目前我的问题是试图在我的链表上打印多个值。

我很确定主要问题来自函数 add_node 但我似乎根本无法改变它。 (有一个 subtract class 选项，但我省略了这个函数，因为它不相关。)

我需要更改什么才能将其添加到链表中？

任何帮助将不胜感激

#include <iostream>
#include <cstring>

struct ToDoList
{
    std::string start_time;
    std::string activity_name;
    int time_for_activity;
    ToDoList *next;
};

void add_node(ToDoList * & head, std::string start, std::string activity,int time )
{
    ToDoList* temp;
    head = new ToDoList;
    temp = new ToDoList;
    head-> start_time = start;
    head-> activity_name = activity;
    head-> time_for_activity = time;
    head-> next = head;
    head = temp;
    temp = temp->next;
    head->next=NULL;
}//in theory this should add another node to the list but it isn't working

int main()
{
    int ans, i = 0;
    std::string start;
    std::string activity;
    int time;
    ToDoList* head;
    ToDoList* a;
    std::cout << "Enter the start time in HH:MM am format: ";
    std::getline(std::cin, start);
    std::cout << "Enter the activity name: ";
    std::getline(std::cin, activity);
    std::cout << "Enter the time for the activity: ";
    std::cin >> time;
    //~ add_node(head);
    add_node(head,start,activity,time);

    std::cout << "\nWelcome to the Linked list menu! What would you like to do? \n0 Add a Node \n1 Display the list \n2 Delete a node \n3 Quit \n";
    std::cin >> ans;
    while(ans != 3)//This should print all the values in the list
    {
        std::cin.ignore();
        if(ans == 0)
        {   
            std::cout << "Enter the start time in HH:MM am format: ";
            std::getline(std::cin, start);
            std::cout << "Enter the activity name: ";
            std::getline(std::cin, activity);
            std::cout << "Enter the time for the activity: ";
            std::cin >> time;
            add_node(head,start,activity,time);

        }
        else if( ans == 1 )
        { 
            a = new ToDoList;//creates new pointer for while loop
            a = head;
            while(a != NULL )//loop used for printing
            {
                std::cout << i << " " << a->start_time << " " << a->activity_name << " " << a->time_for_activity << "\n";
                a = a -> next;
                i++;
            }
            i = 0;//resets integer i
        }
        std::cout << "\nWelcome to the Linked list menu! What would you like to do? \n0 Add a Node \n1 Display the list \n2 Delete a node \n3 Quit \n";
        std::cin >> ans;
    }
    return 0;
}

到目前为止它只打印以下内容:

Enter the start time in HH:MM am format: 10:00 am
Enter the activity name: Office Hours
Enter the time for the activity: 30

Welcome to the Linked list menu! What would you like to do? 
0 Add a Node 
1 Display the list 
2 Delete a node 
3 Quit 
0
Enter the start time in HH:MM am format: 11:00 am 
Enter the activity name: Lunch
Enter the time for the activity: 60

Welcome to the Linked list menu! What would you like to do? 
0 Add a Node 
1 Display the list 
2 Delete a node 
3 Quit 
1
0   0
test

Answer 1

你不需要在每次添加新节点时都在头部创建一个新节点。此外，即使您只是纠正了这一点，每次添加新节点时，您都​​会错误地将 NULL 分配给 head 的下一个，从而阻碍您访问列表中比第一个节点更远的任何成员。最后，您将 temp->next 分配给 temp 是另一个问题来源，它会导致您无法访问除第一个之外的列表元素。

以下是您的代码版本，其中删除了那些不正确的语句。它似乎有效，如您所见here .

void add_node(ToDoList * & head, std::string start, std::string activity,int time )
{
    ToDoList* temp;
    temp = new ToDoList;
    temp-> start_time = start;
    temp-> activity_name = activity;
    temp-> time_for_activity = time;
    temp-> next = head;
    head = temp;
}

此外，虽然它与您的问题没有主要关系，但正如您在我在 Ideone 上试验过的代码版本中看到的那样，我在 main 中将 head 初始化为 NULL，并删除了不必要的动态分配指针 a.需要初始化 head 以便您的某些循环可以正确终止(即与永不终止相反)，同时删除分配对于防止内存泄漏至关重要。