C语言计算双向链表的大小

#include <stdio.h>
#include <stdlib.h>

struct node {
   int data;
   struct node *prev;
   struct node *next;
};

struct node *head = NULL;
struct node *last = NULL;

struct node *current = NULL;

//display the list
void printList() {
   struct node *ptr = head;

   printf("/n[head] <=>");
   //start from the beginning
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->next;
   }

   printf(" [last]/n");
}

//Create Linked List
void insert(int data) {
   // Allocate memory for new node;
   struct node *link = (struct node*) malloc(sizeof(struct node));

   link->data = data;
   link->prev = NULL;
   link->next = NULL;

   // If head is empty, create new list
   if(head==NULL) {
      head = link;
      return;
   }

   current = head;

   // move to the end of the list
   while(current->next!=NULL)
      current = current->next;

   // Insert link at the end of the list
   current->next = link;
   last = link;
   link->prev = current;
}

void sizeof_list() {
   int size = 0;
   struct node *link;

   if(head==NULL) {
      printf("/n Size of list is %d/n", size);
      return;
   }

   link = head;            
   while(link != NULL) {
      link = link->next;
      size++;
   }

   printf("/n Size of list is %d/n", size);
}

int main() {
   insert(10);
   insert(20);
   insert(30);
   insert(1);
   insert(40);
   insert(56); 

   printList();
   sizeof_list();

   return 0;
}