How can my LINKED_LIST pointer still be NULL?

Question

main function:

int main()
{
    if((ban_file = open_file("banned_ips.txt")) == NULL)
        goto exit;

    ban_lst = NULL;
    cpy_to_list(ban_file, ban_lst);
    close_file(ban_file);
    dealloc_list(ban_lst);
exit:
    return 0;
}

cpy_to_list function:

void cpy_to_list(FILE *file, LINKED_LIST *lst)
{
    char *line = malloc(1024);

    while((line = fgets(line, 1024, file)) != NULL)
    {
        add_node(line, lst);
    }

    free(line);
}

add_node, dealloc_list & create_list:

LINKED_LIST *create_list(void)
{
    LINKED_LIST *tmp;

    if((tmp = malloc(sizeof(LINKED_LIST))) == NULL)
        perror("Error during memory allocation");

    return tmp;
}

void add_node(const char *str, LINKED_LIST *lst)
{
    struct list_node *tmp_node;

    tmp_node = create_list();
    tmp_node->str = str;

    if(lst != NULL)
    {
        tmp_node->next = lst;
        lst = tmp_node;
    }
    else
    {
        lst = tmp_node;
        lst->next = NULL;
    }
}

void dealloc_list(LINKED_LIST *ptr)
{
    free(ptr);
}

How come ban_lst is a NULL pointer?

cnicutar · Accepted Answer · 2012-02-07 20:04:26Z

3

cpy_to_list(ban_file, ban_lst);

Whatever it does, this function cannot change what ban_lst points to. Whatever it does, at the end of the call ban_lst will point to where it pointed before. Perhaps you want cpy_to_list to accept a LINKED_LIST ** ?

Your add_node function expresses the problem even better.

lst = tmp_node;

This line only changes the functions idea of the pointer. It doesn't change anything for the caller.

This C FAQ touches on the subject.

answered Feb 7, 2012 at 20:04

cnicutar

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Comments

Carl Norum · Accepted Answer · 2012-02-07 20:04:52Z

3

C is a pass-by-value language. You pass ban_lst to cpy_to_list(), but it can't change it from the caller's point of view.

answered Feb 7, 2012 at 20:04

Carl Norum

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