Question

我正在学习C，而我正试图理解一个更好的链接列表概念。我有一个代码作为一个例子（我把它改成了我想要更好理解的那个）。代码没有任何错误，但再次像我说的那样更好地理解它。请先查看代码，然后在代码下方查看我的问题：

以下是我的struct和我的第一个函数的代码：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

struct client{
char first_name[20];
char last_name[20];
char id;
char reference;
char deposit;
struct client *next; // pointer to the next entry in the clients list
};
typedef struct client new_Client;

/***********************************************************
 * find_Client
 * Finds a place in the list to insert new user
 * Input: User's first and last name, id, reference, deposit
 * Returns: A pointer to teh struct which AFTER we'll insert
 the new struct. NULL if head of the linked-list
 ***********************************************************/

new_Client *find_Client(char *new_last_name, new_Client *head)
 {
    new_Client *prev = NULL; // previous has nothing
    new_Client *curr = head; // current is the head

    while (curr != NULL) //// while the current is not NULL because if it's null it's not head
     {  
      if (strcmp(new_last_name, curr->last_name)<0)
        break;

         prev = curr;
         curr = curr->next;  // pointing to the next node
     }
   return prev;
 }

这是我的下一个功能，我通常会遇到问题

/***********************************************************
 * add_new_Client
 * Adds a new client
 * Input: clients details and the address of the list's head
 * Returns: nothing
 ***********************************************************/

void add_new_Client(char *first_n, char *last_n, char *ident, char *ref, char *dep, new_Client **head)
{
    new_Client *after;

    // create new entry
    new_Client *new_entry = (new_Client*)malloc(sizeof(new_Client));    //allocating size of the struct

    if (NULL == new_entry) 
    {
    exit(1); //Q1: does it mean that it terminates?

    }
    else 
    {
    strcpy(new_entry->first_name, first_n);
    strcpy(new_entry->last_name, last_n);
    strcpy(&new_entry->id, ident);
    strcpy(&new_entry->reference, ref);
    strcpy(&new_entry->deposit, dep);
    }
    after = find_Client(new_entry->last_name, *head); // Q2: don't understand this line..
      if (NULL == after) // new head
        {
            if (NULL == *head) // Q3: for adding the first item (when the head is NULL) --> why is this condition necessary
            {
            new_entry->next = NULL;
            }
           else
           {
               new_entry->next = *head;
           }
           *head = new_entry;
       }else
         { //add in the middle
           new_entry->next = after->next;
           after->next = new_entry;
       }

   }

好的，我把我的问题放在代码中。它们被标记为Q1，Q2和Q3。如果您希望我在我的帖子上编辑并更改提问方法，请告诉我更方便您理解我的问题。

Answer 1

//Q1: does it mean that it terminates?
exit(1);

是的，程序将以1 which is interpreted as failure退出。

// Q2: don't understand this line..
after = find_Client(new_entry->last_name, *head);

此处，after是指向struct new_Client的指针。函数find_Client返回指向使用new_entry->last_name搜索的客户端的指针，并传递head以便为搜索提供开始，某处开始。

// Q3: for adding the first item (when the head is NULL) --> why is this condition necessary
if (NULL == *head)

因为当列表中只有一个元素时，它就是单独的，也就是说，它没有next元素。但是，如果列表大小大于1，则新节点的next元素将指向头部，而此新节点将是新head。也就是说，新节点的next元素将具有前一个head元素的值。这就是为什么这种区别是必要的。

更好地理解链表

1 个答案: