在另一个链接列表(C编程)中包含的链接列表中添加/更新信息

Question

我是一个初学者，试图将信息添加到/更新信息中包含在另一个链接列表中的链接列表中。我的程序是一个简单的联系人列表，其中用户可以将联系人添加到列表中，然后将信息添加到每个联系人（使用单独的信息列表）中。我的附加信息功能会遇到细分错误，而且我看不到问题是什么。以下是结构的定义：

typedef struct info {
  char *name;
  char *value;
  struct info *next;
} Info;
typedef struct contact {
  char *name;
  Info *information;
  struct contact *next;
} Contact;

这是我的添加信息功能：

void addInformation(Contact *myContacts, char *contactName, char *infoName, char *infoValue) {
Contact *ptr = myContacts;
Info *ptr2 = ptr->information;
if (ptr == NULL) {
  printf("Error: No contacts added.n");
  return;
}
while (ptr != NULL) {
  if (ptr->name != contactName) {
    printf("Error: Contact does not existn");
    return;
  }
   else {
      ptr2->name = infoName;
      ptr2->value = infoValue;
      ptr2->next = NULL;
      ptr->information = ptr2;
    }
    ptr = ptr->next;
  }
 return;
}

如果有人可以告诉我我在代码中做错了什么，那将不胜感激！谢谢。

Answer 1

我的第一个反应是，没有分配联系 ->信息的内存。因此，ptr2为空或垃圾。@yano已经提到了。

我试图在原始代码上写评论，然后决定用一些注释编写新版本会更清楚。将代码分为几个函数会更好。

void addInformation(Contact *myContacts, char *contactName, char *infoName, char *infoValue) {
    Contact *ptr = myContacts;
    if (ptr == NULL) {
        printf("Error: bad pointer to contacts storage.n");
        return;
    }
    while (ptr != NULL) {
        if (strcmp(ptr->name,contactName) == 0) {
            // if it's a contact with the name we are looking for
            // then allocate and put info structure as a head of the list of info
            Info *newInfo = malloc(sizeof(Info));
            newInfo->name = infoName;
            newInfo->value = infoValue;
            newInfo->next = ptr->information;
            ptr->information = newInfo;
            return; // get out as we found our contact
        } else if (ptr->next == NULL ) {
            // there is nothing next, we are at the end of the list
            // allocate memory for both contact and information, fill it in
            // add the contact as the last element in the list of contacts.
            Contact* newContact = malloc(sizeof(Contact));
            newContact->name = contactName;
            newContact->information = malloc(sizeof(Info));
            newContact->information->name = infoName;
            newContact->information->value = infoValue;
            newContact->information->next = NULL;
            newContact->next = NULL;
            ptr->next = newContact;
            return; // get out as there is no reason to iterate more, it's the last element in the list
        } else {
            ptr = ptr->next;
        }
    }

    return;
}
int main(int argc, const char * argv[]) {
    // Has to allocate the first contact. Otherwise the pointer management
    // will have to be different and addInformation function has to accept
    // pointer to pointer.
    Info info = { "nn", "vv", NULL };
    Contact myContacts = { "contact 1", &info, NULL };
    addInformation(&myContacts, "contact 2", "info 1", "val 2");
    addInformation(&myContacts, "contact 2", "info 3", "val 4");
    addInformation(&myContacts, "contact 1", "info 5", "val 6");
    printf("end");
    return 0;
}