c-使用双指针访问链表的节点

Question

我正在尝试在链表中添加节点，传递指针的指针，访问成员（*list）->接下来会出现分段错误。有人能指出错误吗，提前谢谢。代码是

void initializeList (node ** head, int data)
{
    *head = malloc (sizeof (node));
    (*head)-> member = data;
    (*head)-> next = NULL;
}
void addNode(node ** list, int data)
{
    node * newNode = NULL;
    newNode = malloc (sizeof (node));
    while (*list->next!= NULL) 
    {
        *list= (*list) -> next;
    }
    newNode -> member = data;
    newNode -> next = NULL;
    (*list) -> next = newNode;
}
void main ()
{
    node * head = NULL;
    initializeList (&head, 5);
    addNode(&head, 6);
}

Answer 1

您不应该取消引用NULL指针，否则会出现Seg Fault。

问题是：

while (*list != NULL)
    {
    *list= (*list) -> next;
    }
newNode -> member = data;
newNode -> next = NULL;
(*list) -> next = newNode;

while循环后*list为NULL
您正在执行(*list) -> next = newNode;

Answer 2

newNode在循环后指向NULL不能设置其成员和下一个值使用while（*list->next！=NULL）

这可以确保您不会取消引用NULL