二叉树节点的 C 声明结构

Question

struct node * newNode(int data)

Can anybody give me an insight what is struct node* ? what does node* is representing?

The intention behind asking this question is :

int main(void){
  struct node *node1,*node2;
  node1 =(struct node*)malloc(sizeof(*node1));
  node2=(struct node*)malloc(sizeof(*node2));
  node1->value=7;
  node1->left=node2;
  node1->right=NULL;
  node2->value=9;
  node2->left=NULL;
  node2->right=NULL;
  printf("%d",node1->left->value);
}

This is working fine ! now here why struct node* is casted? Doing struct *node is showing error.

Answer 1

Seems like you are doing something related to binary tree. First Of all, instead of these 2 lines :-

  node1 =(struct node*)malloc(sizeof(*node1));
  node2=(struct node*)malloc(sizeof(*node2));

Use these (recommended):-

  node1=(struct node*)malloc(sizeof(struct node));
  node2=(struct node*)malloc(sizeof(struct node));

Also, the malloc returns a void pointer and it must be casted into the data type on the Left Hand Side.

node1= malloc( ...);

On the left hand side (above statement) the data type of node1 is struct node and so its pointer type will be struct node * . Therefore, struct node* is used as a cast as below.

node1=(struct node*)malloc( ... );

Answer 2

struct NODE {
   char data[12];
   struct NODE* newNode;    //reference pointer
};

Here NODE* is just a reference structure pointer inside the structure, which can be used to hold the address of the variable declared inside the structure.

Don't get confused : *node :used as pointer 
          where as : node* :is used as reference

  struct NODE{
   char data[12];
   char *next;
};

typedef struct DATA Sample;

Sample *hptr = NULL;