C错误: expression must have arithmetic or pointer type

Question

typedef struct node
{
    Record data;
    struct node *next;
}Node;

Node *head = NULL;

void addRecord(Record x)
{
    Node *previousNode = NULL;
    Node *newNode;
    Node *n;

newNode = (Node*)malloc(sizeof(Node));
newNode->data = x;
newNode->next = NULL;

if (head == NULL)  // The list is empty
{
    head = newNode;
}
else       // The list is not empty
{
    n = head;
    while (n->next != NULL)
    {
        ***if (n->data < newNode->data && n->next->data > newNode->data)*** // Insertion Sort
        {
            // We have to put it between these 2 nodes
            newNode->next = n->next;
            n->next = newNode;
            return;
        }
        else
        {
            previousNode = n;
            n = n->next;
        }
    }
    n->next = newNode;
}

}

我在插入排序的 if 函数内的代码中遇到此错误。该程序表示“n”必须具有算术或指针类型。问题出在哪里？

Answer 1

C 中不支持运算符重载，因此您无法使用 > 运算符比较 Record，除非将其 typedef 编辑为 int 或其他算术或指针类型。

要比较结构之类的东西，请定义比较函数并使用它。

示例:

typedef struct {
    int a, b;
} Record;

/*
return positive value if *x > *y
return negative value if *x < *y
return 0 if *x == *y
*/
int cmpRecord(const Record* x, const Record* y) {
    if (x->a + x->b > y->a + y->b) return 1;
    if (x->a + x->b < y->a + y->b) return -1;
    return 0;
}

/* ... */
    while (n->next != NULL)
    {
       if (cmpRecord(&n->data, &newNode->data) < 0 && cmpRecord(&n->next->data, &newNode->data) > 0) // Insertion Sort
        {
/* ... */