How to Use LLIST *mylist[N];

Question

I get this:

LLIST *mylist[N];

Where N is the number of rows of the input file. Then mylist[i] is a pointer to the ith linked list.

I call a function

LLIST *list_add(LLIST **p, int i){
    LLIST *n;

    n = (LLIST *) malloc(sizeof(LLIST));

    if (n == NULL)
        return NULL;

    n->next = *p; /* the previous element (*p) now becomes the "next" element */
    *p = n;       /* add new empty element to the front (head) of the list */
    n->data = i;
    return p;
}

So in my Main I say something like

LLIST *mylist[N];
list_add(&mylist[0],1);
list_add(&mylist[0],2);
list_add(&mylist[1],3)
list_add(&mylist[1],4);
list_print(mylist[0]); // Print mylist[0]
list_print(mylist[1]); // Print mylist[1]

My Print_list function is:

Void *list_print(LLIST *n) {
    if (n == NULL){ 
        printf("list is empty\n");
    }

    while (n != NULL){
        printf("%d",n->data); 
        n = n->next;
    }
}

When I do list_print(mylist[0]), it prints out 2 1.

When I do list_print(mylist[1]), I get a segmentation fault.

What is going on?

Answer 1

You don't seem to initialize the contents of the mylist array, and it might contain values that point at random memory locations. When just adding new elements this doesn't matter, new elements are inserted at the beginning, never accessing the invalid pointers. But when you try to print the list, the print function will follow the invalid pointer at the end of the list and produce a segmentation fault.

To avoid this add an initializer to the array declaration:

LLIST *mylist[N] = {NULL};

This will set all elements of mylist to NULL.

Answer 2

This does work LLIST *mylist[10] = {NULL}; But would if I wanted to do this I get errors:

x=10; LLIST *mylist[x] = {NULL};