c - 我第一次使用链表。想要一些关于如何解决某些问题的批评和建议

Question

该程序适用于酒店洗衣服务。用户输入他们的房间号、名字和姓氏，以及他们要洗的元素数量。该信息被放入链表中的节点中。在主菜单中，用户可以添加更多房间请求、更新请求、打印请求或退出程序。

结构可以定义为

struct request{
    int room_number;
    char first[NAME_LEN+1];
    char last[NAME_LEN+1];
    int num_items;
    struct request *next;
};

我的函数遇到了一些问题:

1. 追加函数:

/*
APPEND FUNCTION:
Gets the room number, first name, last name, and the number of items the user wants to wash.
Creates a new node and appends it to the end of the linked list.
*/
struct request *append_to_list(struct request *list)
{
 struct request *new_node, *last_node, *search;


    new_node = malloc(sizeof(struct request));
    //new_node->next = NULL;

    if(new_node == NULL)
    {
        printf("Error allocating memory.\n");
        return list;
    }

    //get room number
    printf("Enter room number: ");
    scanf("%d", &new_node->room_number);

    //search to see if the room number already exists in the LL.
    for(search = list; search != NULL; search = search->next)
    {
        if(search->room_number == new_node->room_number)
        {
            printf("Room request already exists. Update request using main menu.");
            return list;
        }
    }

    //get first and last name
     printf("Enter first name: ");
    read_line(new_node->first, NAME_LEN+1);
    printf("Enter last name: ");
    read_line(new_node->last, NAME_LEN+1);

    //get the number of items.
    printf("Enter the number of items you wish to wash: ");
    scanf("%d", &new_node->num_items);

    new_node->next = list;


    //if list is empty, return pointer to newly created linked list.
    if(list == NULL)
    {
        list = new_node;
        return new_node;
    }
    //else add request to the end of the LL and return pointer to the LL.
    else
    {
        last_node = list;
        while(last_node->next!=NULL)
            last_node = last_node->next;
    }
    last_node->next = new_node;


 return list;

}

我遇到的一些问题是由于某种原因我不能发出超过两个请求。我收到一个错误，程序崩溃了。

1. 更新函数:

/*
UPDATE FUNCTION:
User enters their room number and the node containing the room number is updated with the number of items the user wants to add on.
*/
void update(struct request *list)
{


    struct request *search;
    int add_items;

    //ask to enter room num
    printf("Enter room number: ");
    int room;
    scanf("%d\n", &room);

    //find matching room num
    for(search = list; search != NULL; search = search->next)
    {

        if(search->room_number == room)
        {
            //ask to enter num of items to be added and update num of items


            printf("How many items would you like to add: ");
            scanf("%d\n", &add_items);
            search->num_items = search->num_items + add_items;
            search = search->next;
            return;

        }
    }
    //if room num is not found, print a message.
            printf("Could not find request.");
    return;
}

我在使用此功能时遇到的一个问题是，当我输入房间号时程序会停止...它不会崩溃，它似乎卡住了...不太清楚为什么。

1. 最后是打印函数:

/*
PRINTLIST FUNCTION:
Prints all the nodes in list.
*/
void printList(struct request *list)
{
  //print room num, first and last name, and num of items for all requests on the list.
    while(list != NULL)
    {
        printf("%d ", list->room_number);
        printf("%s ", list->first);
        printf("%s ", list->last);
        printf("%d\n ", list->num_items);
        list = list->next;
    }
}

我对这个函数的唯一问题是它会不停地无限打印所有节点。

感谢任何帮助。谢谢!

Answer 1

1. 您需要决定是要将 new_node 放在列表的末尾还是开头。 list = new_node 将它放在开头，后续循环将它放在末尾，因此您创建了一个没有结尾的循环列表，并且您的下一个插入操作陷入了无限循环。如果你想让new_node在开头，你不需要搜索结尾。如果你想在最后，那么 new_node->next 必须设置为 NULL，而不是 list。
2. 您的 for 循环体只执行一次，因为您在 if 语句的两个分支中都有 return。
3. 这可能是因为上面的要点 1。