Acessing another structure's member through a pointer

Question

#include<stdio.h>

int main()
{
    struct node
    {
        int data;
        struct node *next;

    };

    struct node n1,n2,n3;
    int i;
    n1.data = 100;
    n2.data = 200;
    n3.data = 300;

    n1.next = &n2;
    n2.next = &n3;

    i = n1.(*next).data;
    printf("%i\n",i);

    printf("%i\n", n2.next->data);

    return 0;
}

My Question is simple. How come i am not able to access n2 node via n1 node when i use this i = n1.(*next).data; I get an error. However if i change it to i=n1.next->data

I thought (*x).y and x->y mean the same thing.

Answer 1

You're right, (*x).y and x->y mean the same thing, that's why you should write (*n1.next).data (equivalent to n1.next->data as n1.next is your x).

EDIT: The parenthesis are so due to the evaluation order:

1. Take n1: n1 (a struct node)
2. Take member next: (n1).next (a struct node *)
3. Deference the pointer: *((n1).next) (a struct node)
4. Take member data: (*((n1).next)).data (an int)

Read what the C11 standard says in the @Christophe answer, and notice how the . operator used in steps 2 and 4 above where used with struct node as first operands (see steps 1 and 3) and member names as seconds operands.