来自不兼容指针类型的赋值[默认启用]

Question

为什么我在insertNode() 中收到此警告：

warning: assignment from incompatible pointer type [enabled by default]|

在这一行：

 head->next = newNode; //point head's next to the newNode

和

warning: initialization from incompatible pointer type [enabled by default]|

在这一行：

 Node *current = head->next; 

在我的main() 函数中，我也有这个警告：

warning: passing argument 1 of 'insertNode' from incompatible pointer type [enabled by default]|

在这一行：

insertNode(&head, num);

我的代码中有许多与这些警告类似的其他警告。我该如何修复它们？

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

void insertNode(Node *head, int data){
    Node *newNode = (Node *)malloc(sizeof(Node));
    newNode->data = data;

    if(head->next == NULL){
        head->next = newNode; 
        newNode->next = NULL;
    }

    else{
        Node *current = head->next; 
        while(current != NULL && current->data < data){
            current = current->next;
        }
        newNode->next = current->next;
        current->next = newNode; 
    }
}

int main(int argc, char *argv[])
{
    Node *head = malloc(sizeof(NodeStruct));
    head->next = null;
    insert(head, 22);
    insert(head, 55);
    insert(head, 44);
    insert(head, 2);
    insert(head, 2112);
    insert(head, 3);


    printList(head);
    return 0;

}

Answer 1

在main-

insertNode(&head, num);
           ^ don't pass address

它需要Node *，你应该这样称呼它-

insertNode(head, num);

其他警告是由于上述错误，因为您将head 的地址而不是head 传递给函数。

也可以在struct NodeStruct 更改这个 -

struct Node *next;         // you cannot reference Node in struct itself

到-

struct NodeStruct *next;   //You need to use structure name