Thread: 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

Or a union.

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

Cheers WhiteFlags.