删除链表中的第一个节点

问题描述 投票:0回答:1

我试图删除链表中的第一个节点,我不知道如何去做。我的链表看起来像这样

typedef struct availableForRent{
   int milage;
   char plateNum[8];
   struct availableForRent * next;
} availablreForRent;

我的链表的初始节点是硬编码的

struct availableForRent * head = NULL;
head = malloc(sizeof(struct availableForRent));
head->milage = 190000;
fillString(head->plateNum);
head->next = NULL;

fillString只获取用户输入并将其放入数组中

我通过这样的推送功能将一个成员添加到我的列表中

void pushAvailable(struct availableForRent * head) {
   struct availableForRent * current = head;
   while (current->next != NULL) {
      current = current->next;
   }

   current->next = malloc(sizeof(struct availableForRent));
   printf("Enter a milage amount: ");
   scanf("%d", &current->next->milage);
   fillString(current->next->plateNum);
   current->next->next = NULL;
}

我删除第一个成员的功能看起来像这样

struct availableForRent * next_node = *head;

if (next_node->next == NULL) {
   printf("Cannot remove member as it is the only data in the list!\n");
   return;
}

next_node = next_node->next;
free(*head);
*head = next_node;

当我运行我的程序时,我收到此错误消息,

rentalQ1(2799,0x7fff9d09b380) malloc: *** error for object 0x7ffee8f62a08: pointer being freed was not allocated
*** set a breakpoint in malloc_error_break to debug
Abort trap: 6

这是否意味着我试图释放的节点不存在或其他什么?

c memory linked-list
1个回答
0
投票

似乎没有mallocing next_node是我的问题。

next_node = malloc(sizeof(struct availableForRent));
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