CODEWITHSUNDEEP
c++pointers
user5005768Himadree
user5005768Himadree

Reputation: 1427

How to get the address of node->next pointer

I was writing code on Doubly Linked List for the following function 

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

/* Given a node as next_node, insert a new node before the given node */
void insertBefore(struct node** next_node, int new_data)

I called the insertBefore() from the main function:

 insertBefore(&head, 2);  // it works fine

But i can't do the same fore head->next, because next pointer is just node* Not node**

insertBefore(head->(&next), 2);  // it doesn't works

How can i solve the problem.please help.Thanks

Upvotes: 0

Views: 4043

Answers (4)

user2672107
user2672107

Reputation:

Change your function to take a reference to the node pointer.

void insertBefore(node*& next_node, int new_data);
...
insertBefore(head->next, 2);

Upvotes: 1

Error - Syntactical Remorse
Error - Syntactical Remorse

Reputation: 7887

It seems illogical to dereference a pointer than reference its address...

&(head->next) will work but it works because you dereference the pointer (grab the value that the pointer points to) than grab the address of that pointer again using &.

-> is the same as .(*next). The arrow is just the easiest way to get what the pointer points to. If you want the value of the pointer itself all you have to do is head.next.

Upvotes: 0

sameerkn
sameerkn

Reputation: 2259

head is a pointer in case of insertBefore(&head, 2);

next is also a pointer and a member of struct node.

So, if you want to access next via head (i.e pointer to some node) then,

syntax would be head->next. Now you can get address of next by &(head->next).

Upvotes: 1

nvoigt
nvoigt

Reputation: 77304

Assuming your function works, &(head->next) is what you need. 

insertBefore(&(head->next), 2);

Upvotes: 1

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