LinkedList Pointer (REPOST - diff version)

[r.devaraj@gmail.com]

Some typo and confusion in previous post
(http://groups.google.com/group/comp....3fdc9a2d49e9ac)
made it difficult for me to fetch correct answers.
Here's the original problem.

Consider Linked List

struct node{
int data;
struct node * next;
}list;


No problem until you are dealing with a pointer variable(list).
there, you can access[color=blue][color=green]
>> memory where it is pointing to.
>> own address of pointer variable.[/color][/color]

but think in case &(list->next)[color=blue][color=green]
>>you can just access the target.
>> no clue about (atlest for me, now) what pointer comes to picture[/color][/color]
or where it is stored.,


Description:

Consider foll. routines.

a function to add a node in the list

void Push(struct node** headRef, int newData);
Given an int and a reference to the head pointer (i.e. a struct
node** pointer to the head pointer), add a new node at the head of the
list with the standard 3-step-link-in: create the new node, set its
..next to point to the current head, and finally change the head to
point to the new node.


and a routine using it is,


void BasicsCaller() {
struct node* head;
int len;
head = BuildOneTwoThre e(); // Start with {1, 2, 3}
Push(&head, 13); // Push 13 on the front,
// yielding {13, 1, 2, 3}
// (The '&' is because head is
//passed
// as a reference pointer.)
Push(&(head->next), 42); // Push 42 into the second
//position
// yielding {13, 42, 1, 2, 3}
// Demonstrates a use of '&' on
// the .next field of a node.
// (See technique #2 below.)
}



First call for push(&head,13) is obvious and works fine.


Now this is the single question:
Just explain what happens in the call Push(&(head->next), 42)
what is passed to Push and what it is done in push.


if you couldn't get me completely,

http://cslibrary.stanford.edu/105/LinkedListProblems.pdf

Thanks.
Deva.

[manoj1978@gmail.com]

Re: LinkedList Pointer (REPOST - diff version)


r.devaraj@gmail .com wrote:[color=blue]
> Some typo and confusion in previous post
> (http://groups.google.com/group/comp....3fdc9a2d49e9ac)
> made it difficult for me to fetch correct answers.
> Here's the original problem.
>
> Consider Linked List
>
> struct node{
> int data;
> struct node * next;
> }list;
>
>
> No problem until you are dealing with a pointer variable(list).
> there, you can access[color=green][color=darkred]
> >> memory where it is pointing to.
> >> own address of pointer variable.[/color][/color]
>
> but think in case &(list->next)[color=green][color=darkred]
> >>you can just access the target.
> >> no clue about (atlest for me, now) what pointer comes to picture[/color][/color]
> or where it is stored.,[/color]

It contains the address of list-> next.

that is address of the field next,inside the structure pointed by list.
[color=blue]
>
>
> Description:
>
> Consider foll. routines.
>
> a function to add a node in the list
>
> void Push(struct node** headRef, int newData);
> Given an int and a reference to the head pointer (i.e. a struct
> node** pointer to the head pointer), add a new node at the head of the
> list with the standard 3-step-link-in: create the new node, set its
> .next to point to the current head, and finally change the head to
> point to the new node.
>
>
> and a routine using it is,
>
>
> void BasicsCaller() {
> struct node* head;
> int len;
> head = BuildOneTwoThre e(); // Start with {1, 2, 3}
> Push(&head, 13); // Push 13 on the front,
> // yielding {13, 1, 2, 3}
> // (The '&' is because head is
> //passed
> // as a reference pointer.)
> Push(&(head->next), 42); // Push 42 into the second
> //position
> // yielding {13, 42, 1, 2, 3}
> // Demonstrates a use of '&' on
> // the .next field of a node.
> // (See technique #2 below.)
> }
>
>
>
> First call for push(&head,13) is obvious and works fine.
>
>
> Now this is the single question:
> Just explain what happens in the call Push(&(head->next), 42)
> what is passed to Push and what it is done in push.[/color]

It will pass the address of next in the structure pointed by head.
Push will allocate a node,put 42 inside that.and store address of that
node in this address.

So after returning from push,head->next will contain address of newly
allocated node.
[color=blue]
>
>
> if you couldn't get me completely,
> http://cslibrary.stanford.edu/105/Li...stProblems.pdf
>
> Thanks.
> Deva.[/color]