[C]: When a pointer to a pointer to a struct is passed in a function, why can't it change the elements of that struct? [duplicate]答案

【问题标题】：[C]: When a pointer to a pointer to a struct is passed in a function, why can't it change the elements of that struct? [duplicate][C]: When a pointer to a pointer to a struct is passed in a function, why can't it change the elements of that struct? [duplicate]
【发布时间】：2017-12-13 18:53:17
【问题描述】：

struct Node {
int data;
struct Node *next;
};

void insertAfterHead(struct Node **a, int x) {
struct Node *newNode = (struct Node *)malloc(sizeof(struct Node));
newNode -> data = x;
newNode -> next = *a -> next;
*a -> next = newNode;
}


int main() {
    struct Node *head = NULL;
    struct Node *second = NULL;
    struct Node *third = NULL;

    head = malloc(sizeof(struct Node));
    second = malloc(sizeof(struct Node));
    third = malloc(sizeof(struct Node));

    head -> data = 5;
    head -> next = second;

    second -> data = 10;
    second -> next = third;

    third -> data = 15;
    third -> next = NULL;

    insertAfterHead(&head, 10);

    return 0;
}

这是我在运行代码时收到的错误消息：

错误：成员引用基类型'struct Node *'不是结构或联合新节点 -> 下一个 = *a -> 下一个；

错误：成员引用基类型'struct Node *'不是结构或联合 *a -> 下一个 = 新节点；

请尽可能描述，因为我是 C 初学者 :)

【问题讨论】：

错误信息与代码不对应。并将其粘贴到此处，不是作为图像，而是作为文本。
请将您的输出（错误消息）作为文本发布，并请发布您从您发布的代码（而不是其他代码）中获得的错误消息。
在这种情况下，不需要传递指向指针的指针。
@EugeneSh。抱歉，我在询问时重命名了变量（为了澄清）。也添加了错误消息。
@JohnnyMopp - 是的，但我忘记了我想问这个问题的上下文，所以我编了：P

标签： c pointers struct linked-list

【解决方案1】：

*headRef->next

意思

*(headRef->next)

但你想要

(*headRef)->next

【讨论】：