/* File:     dyn_linked_list.c
 * Purpose:  Illustrate how the linked list 5 -> 9 -> 7 can be built
 *           using pointers and malloc
 *
 * Compile:  gcc -g -Wall -o dyn_linked_list dyn_linked_list.c
 * Run:      ./dyn_linked_list
 *
 * Input:    None
 * Output:   The ints in the list
 */
#include <stdio.h>
#include <stdlib.h>

struct list_node_s {
   int data;
   struct list_node_s* next_p;
};

void Print(struct list_node_s* head_p);

int main(void) {
   struct list_node_s* head_p;
   struct list_node_s* temp_p;

   temp_p = malloc(sizeof(struct list_node_s));
   temp_p->data = 7;
   temp_p->next_p = NULL;
   head_p = temp_p;

   temp_p = malloc(sizeof(struct list_node_s));
   temp_p->data = 9;
   temp_p->next_p = head_p;
   head_p = temp_p;

   temp_p = malloc(sizeof(struct list_node_s));
   temp_p->data = 5;
   temp_p->next_p = head_p;
   head_p = temp_p;

   Print(head_p);

   return 0;
}  /* main */



/*-----------------------------------------------------------------
 * Function:   Print
 * Purpose:    Print list on a single line of stdout
 * Input arg:  head_p
 */
void Print(struct list_node_s* head_p) {
   struct list_node_s* curr_p = head_p;

   printf("list = ");
   while (curr_p != NULL) {
      printf("%d ", curr_p->data);
      curr_p = curr_p->next_p;
   }  
   printf("\n");
}  /* Print */