Difference between . and ->

Hi there,

I am just trying to learn how to make linked lists and have made the following function:

Code:

---

list* addList()  {
    list* firstList = NULL;
        list* newList = malloc(sizeof(list));
        newList -> count = 128;
        newList -> next = firstList;
       
        return newList;
}

---

This works fine and when I have the following printf statement:

Code:

---

list* bigList = addList();
printf("%d\n", bigList -> count);

---

then 128 gets printed out, as I hoped it would.

My question is that when I try the following printf instead:

Code:

---

printf("%d\n", bigList.count);

---

then I get an error saying that a struct is required. As I understand it, my addList() function returns a pointer to the list element labeled newList in the function. This list element has a member called count. So why is it that I can't use the membership operator '.' in this case? And what is the essential difference between '.' and '->'?

Many thanks in advance.

Joe

-> requires pointer to the structure on it's left side, whereas the '.' operator requires a structure variable on it's left side not a pointer.
Eg.

Code:

---

struct book
{
int page;
char name;
float price}b1;
b1.page //correct
b1->price //incorrect

---

Or a union.

Excellent. Thank you Ben10 for the explanation. I understand now. :cool:

Cheers WhiteFlags.