Using the same pointer in a linked list

Question

I am trying to understand so called linked list in c using struct-nodes. What I want is to use the same pointer through the code but when inplementing this the for loop prints the last character. I think I understand why - probably because the pointer points at the second and the last node in the program.

The program puts two characters in to succesive nodes - the letters "h" and "e". And the purpose is to print these characters.

So the question - How do I go back to the first node using this pointer? Is the only answer to create two pointer - i.e. one pointer called for example startpointer and another pointer called nextpointer? Or could I somehow use one and same pointer throug the code?

thanks!

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

 struct listNode {
    char c[2];
    struct listNode *ptrNextNode;
 };

 typedef struct listNode ListNode;
 typedef ListNode *ListNodePtr;


 int main(void) {

    ListNode list_node;

    ListNodePtr list_node_ptr;

    list_node_ptr = malloc(sizeof(list_node));

    strcpy(list_node_ptr->c, "h");

    list_node_ptr = list_node_ptr->ptrNextNode;

    list_node_ptr = malloc(sizeof(list_node));

    strcpy(list_node_ptr->c, "e");

    int i;
    for (i = 0; i < 2; i++) {
     printf("%s", list_node_ptr->c);
    }

 return EXIT_SUCCESS;
} 

Answer 1

The road to the dark side comes with this assignment:

list_node_ptr = list_node_ptr->ptrNextNode;

Here you overwrite the original pointer with an uninitialized value. This makes you loose the original list_node_ptr causing a memory leak. And if you, after the above assignment, tried to use list_node_ptr except as a recipient to an assignment, it would have been undefined behavior.

Although, in the very next line, you overwrite the pointer again, making it again a valid pointer, and avoid the undefined behavior.

All in all, you're not making a linked list by the code in the question. You could do something like

ListNodePtr head, list_node_ptr;

head = list_node_ptr = calloc(1, sizeof(*list_node_ptr));
list_node_ptr->c[0] = 'c';
list_node_ptr->next = calloc(1, sizeof(*list_node_ptr));

list_node_ptr = list_node_ptr->next;
list_node_ptr->c[0] = 'e';

for (list_node_ptr = head; list_node_ptr != NULL; list_node_ptr = list_node_ptr->next)
    printf("%s\n", list_node_ptr->c);

In the above code, I allocate the next pointer before reassigning it to list_node_ptr. I also save the original head of the list in the head variable so you can iterate the list.

Answer 2

Yes, you need a doubly linked list, which is to use another pointer to point to the previous node.