Thread: Help Related Linked Lists Basics

Code:

#include<stdio.h>
#include"alloc.h"
struct node
{
int data;
struct node *link;
};
void main()
{
struct node *p,*q,*r;
clrscr();
p=(struct node *)malloc(sizeof(struct node));
p->data=1;
q=malloc(sizeof(struct node));
q->data=2;
p->link=&q->data;
r=malloc(sizeof(struct node));
r->data=3;
q->link=&r->data;
r->link=NULL;
printf("p data=%d p address=%u",p->data,p->link);
printf("\n%d",*p->link->link->link);
printf("\n%d",*p->link->link);
getch();
}

this is a simple piece of the code every thing is clear but i m having prob with p->link->link->link or p->link->link statements i m unable to understand how they are working
i know that *p->link->link will print the value at r and p->link->link->link will print 0 "NULL" ascii but can someone explain actually what is happening internally in these p->link->link....link statements what are these and how these are working THANX in advance.

Assuming that your code worked properly, here is how it would work:

Imagine a bunch of boxes in a row. Between each box is an arrow pointing to the next box. P is the first box in this row, and Q is the box that it points to. Each box holds a number; that's it's value.

Where does this fit in your code? Well, P is your first node*. the value in the box is the value of p->data. And Q is the value that p->link points at. I'm assuming you understand pointers if you are using linked lists. Basically, a linked list points to another instance of the same structure, with the last item in the list pointing at NULL.

So when you say p->link, you're saying q. When you say p->link->link, you're saying q->link, or r. When you say p->link->link->link, you're saying q->link->link, or r->link, or NULL.

Hope that's what you wanted to know.

EDIT: * is the de-reference operator when you use it in *p->link->link, so it will dereference the value of p->link->link. Except that I'm not sure which operator has higher priority, * or ->, so it might dereference p before trying to point to link? Just to be safe, I'd say *(p->link->link) to get an actual node object.

I think the * operator has lower priority, isn't that the whole reason for the -> operator replacing the . operator for pointers.

*looking up my C manual* ok, subscripts[], function calls(), field.selection, indirect->selection, ++(postfix), --(postfix) have the highest precedence. then ++(prefix), --(prefix), sizeof, ~!-+&*(unary prefix) are next. so that means that the *p->link->link will dereference p->link->link

thank you every body who tried to answer my question but specially @vart who cleared my doubt and i didnt read what is written below vart's post but thanx everybody i love this forum