Thread: Linked list using double pointer

I have a doubt about double pointer....I mean pointer to pointer
Here is my code

Code:

#include<stdio.h>
#include<stdlib.h>
struct list
{
       int num;
       struct list * link;
};

typedef struct list node;

void display( node **p )
{
     node **temp = p;
     if( **p != NULL )
     printf("NOT NULL");
}

main()
{
       node *head;
       display(&head);
}

I'm not showing the actual program....This is for clarifying my doubt

...node *head means it holds address of a variable of type node,so that we can access that variable like this *head ..ie. value at address head
now i am passing address of head to the function....so type of argument is pointer to a pointer.
..ie...void display(node**p)
...
what is the problem with the statement if(**p!=NULL)

*p means value at address p....ie address of pointer variable head...**p means value at address address of pointer variable p....ie.pointer variable..am i right?
But compiler says that its type is node...not node *...what is the problem...Please give a good explanation on double pointers

You don't want to pass pointer to pointer... just pass and work with the pointer...

Code:

include<stdio.h>
#include<stdlib.h>
struct list
{
       int num;
       struct list * link;
};

typedef struct list node;

void display( node *p )        // here you are passing in a pointer to your first node
{
     node *temp = p;            // making a copy of the pointer
     if( p != NULL )                 // p is already apointer to a node
     printf("NOT NULL");
}

main()
{
       node *head = NULL;    // head will not be null unless intialized to NULL
       display(head);              // no need for the address of, head is already a pointer
}

Originally Posted by CommonTater

The problem is that you are using a lot of unnecessary pointer to pointer stuff that won't work in your application.

If you already have the pointer you need, making a pointer to it is foolish because you end up with an unitialized pointer to an unitialized pointer.

just *p != NULL is enough??
why??

Originally Posted by linuxlover

just *p != NULL is enough??
why??

Just p != NULL is enough.

node *head is defined as a pointer.

Your function accepts it as a pointer ... void display (node *p)

So inside your function you have the pointer to head...
You can test its value directly.

Originally Posted by CommonTater

Just p != NULL is enough.

node *head is defined as a pointer.

Your function accepts it as a pointer ... void display (node *p)

So inside your function you have the pointer to head...
You can test its value directly.

I know how to pass a pointer to a function and manipulate it.....That's not what i needed....I need to know how to pass address of a pointer to a function and manipulate it.....You should not recursively come out again and again saying the same thing.....

@linuxlover: If you have a declaration like node **p, then p contains an address (of a pointer to a node). *p also contains an address (of a node). **p contains a node. Thus, something like **p != NULL doesn't make sense, since **p is a node (a struct), and structs can't be NULL, only pointers can. This is usually done so you can modify the value of *p (the address of a node) inside a function, for example:

Code:

void new_node(node **head)
{
    *head = malloc(sizeof(**head));
}
...
node *p;
new_node(&p);

I generally avoid this if I can do something like the following instead:

Code:

node *new_node(void)
{
    return malloc(sizeof(node));
}
...
node *p = new_node();

But sometimes you need to return multiple things from a function (maybe an function that initializes several pointers), and you have to pass in a pointer by address.

Thanks I got the idea...ie *p means value at address hold by p ..p holds address of pointer variable......value at that address is head..and i can also right
*p->num=10;
or
**p.num =10;


am i right???