c - 使用链表创建堆栈时出错

Question

我正在使用链表创建一个堆栈(我只在推送部分)。我已将我的结构声明为

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

我有 3 个全局声明的指向 LinkedStack 的指针。

struct LinkedStack *first = NULL;
struct LinkedStack *previous = NULL;
struct LinkedStack *current = NULL;

在main()中我写了这个

int data = 0, choice = 0;
if(current == NULL) {
     printf("\nNo Memory Allocated");
}

printf("\n1. Push Data");
printf("\n2. Pop Data");
printf("\n3. Display The Stack");
printf("\n4. Exit");

scanf("%d", &choice);

switch(choice) {
    case 1:
        printf("Enter the data:: ");
        scanf("%d", &data);
        push(&current, data);
        break;
}

我写的推送函数是这样的-

void push(struct LinkedStruct **s, int usrdata) {
    if(first == NULL) {
        first = *s;
    }  
    if(previous != NULL) {
        previous->next = *s;
    } 

    *s->data = usrdata;
    previous = *s;
    *s->next = NULL;
}

但这会导致编译失败。错误是-

Request for member 'data' in something which isn't a structure or an union
Request for member 'next' in something which isn't a structure or an union

这些错误的行号如下:

  *s->data = usrdata;
  previous = *s;
  *s->next = NULL;

你能告诉我为什么会发生这种情况吗？

Answer 1

“箭头”运算符具有更高的precedence比解引用运算符。因此，从编译器的角度来看，您已经编写了例如

*(s->data)=usrdata;

您需要使用括号:

(*s)->data=usrdata;