/* This program tries to demonstrate two different types of dynamic
 * memory allocation using strings.
 */
#include <stdio.h>
#include <string.h>

typedef struct node {
    char * string;
    struct node *next;
} Node;

/* Insert the new string into the node.  Use only as much space as needed. Treat the linked list like a stack. Note the address of
 * the pointer to the linked list has been passed in as a parameter. This allows it's value to change as required -- it also means
 * that it must be dereferenced when used.
 */
void insert( Node ** head, char * new_string ) {
    Node * ptr;

    ptr = (Node *) malloc ( sizeof(Node) );
    if ( ptr == NULL ) {
	fprintf(stderr,"Error! Could not allocate space\n");
	exit(-2);
    } /* if */

    ptr->next = *head;

    /* One extra space, for the '\0' character implicit in strings, is required. */
    ptr->string = (char *) malloc(sizeof(char)*(strlen(new_string)+1));
    strcpy( ptr->string, new_string );
    
    *head = ptr;
} /* insert */

int main( void ) {
    char buffer[20];
    Node * root = NULL, *tmp;

    printf("Enter a list of words, each under 20 characters in length.\n");
    printf("Press <Ctrl>-d to signal EOF\n");
    for (;;) {
	if ( scanf("%s", buffer) == EOF ) break;
	insert(&root, buffer);	/* Note that "&root" and not "root" was passed in. */
    } /* for */

    printf("The list was, in reverse order:\n");
    for ( tmp = root; tmp != NULL; tmp = tmp->next ) {
	printf("%s\n", tmp->string);
	free( tmp->string );
	free( root );
	root = tmp->next;
    } /* for */
}