double astrix pointer question..

**Elysia** · 12-29-2008

Originally Posted by transgalactic2

so **p point to a Node * type variable

Of course not.
p points to a Node*!
Pointers to pointers are not special!
So dereference your pointer and you get the original Node* variable.
But wait, you can dereference that original variable too, so you get the Node value (or data).

Code:

int* x = malloc( sizeof(int) );
*x = 100;
int** y = &x;
printf("%i", **y);
printf("%i", *x);
// y = address of x (pointer to x!)
// *y = x (the address in x!)
// **y == 100 (the value pointed to by x!)
free(*y);
*y = malloc( sizeof(int) );
**y = 200;
printf("%i", **y);
printf("%i", *x);
free(x);

How does this code work and what does it print, do you think?

**~~transgalactic2~~** · 12-29-2008

Originally Posted by laserlight

Okay, at least you got the first part. Now, what is the type of elt->next? What is the type of *p?

*p is the value of P

elt->next is type node

**Elysia** · 12-29-2008

Originally Posted by transgalactic2

*p is the value of P

Huh?

elt->next is type node

That is incorrect. Read the source again.

**~~transgalactic2~~** · 12-29-2008

i ment the value of p
*p is the value of variable p

**Elysia** · 12-29-2008

Yes. Yes, it is. But what type is *p?

**~~transgalactic2~~** · 12-29-2008

Node *

I got a linked list i dont know what points to what
??

**Elysia** · 12-29-2008

Well then, if you understand how the p2p works, then you should be able to understand the code. Is there a specific part you fail to understand yet?

**~~transgalactic2~~** · 12-29-2008

In this code
I pointed in a comment.
it all comes down to who is the "next" of who
and i cant undertand it here?

Code:

typedef struct node {
    int value;
    struct node *next;
}Node;

void what1(Node ** p,int num){
   Node *elt;
   elt=(Node*)malloc(sizeof(Node));  //create a node called elt ,assign it some memory
   elt->next=*p;     //from this point i dont know what the effect of the double astrix
   elt->value=num;
   *p=elt;
}

**Elysia** · 12-29-2008

It assigns *p to it, and what is *p?

**~~transgalactic2~~** · 12-29-2008

*p is of type Node*

but it doesnt tell me who is the next object of who??

**Elysia** · 12-29-2008

Yes, it does. This line says is all:
*p=elt;
Besides, it is a linked list and at a glance, it should be obvious what it does.

**~~transgalactic2~~** · 12-29-2008

if it were p=elt
i would say it puts the address of elt and puts it to p

but its *p?

**~~transgalactic2~~** · 12-29-2008

i could say
i would say it puts the address of elt and puts it to *p

but it meaningless to me
because its a linked list
i cant see what points to what

**laserlight** · 12-29-2008

The key to the linked list is on this line:

Code:

elt->next=*p;

This says that the node pointed to by elt now has a next pointer that points to the node pointed to by *p. In more graphical terms, this tongue twister is like this:

First, we have two pointers that point to nodes:

Code:

*p --> [node #1]
elt --> [node #2]

Now, we get elt's next pointer to point to what *p is pointing to:

Code:

elt->next --> [node #1]

This means that we now have two nodes linked together:

Code:

[node #2] --> [node #1]

Now, here's the catch. elt points to node #2, but elt itself is a local variable. When the function ends, elt will be destroyed. What to do?

We can choose to return elt, but in this case we choose to make *p point to what elt points to. As such, the caller now has a pointer to node #2.

**itCbitC** · 12-29-2008

Originally Posted by transgalactic2

if it were p=elt
i would say it puts the address of elt and puts it to p

but its *p?

That's because they are of the same type, both elt and *p are pointers to type Node that's why they can be assigned.
You can't equate p = elt because they are of different types; p is a pointer to another pointer while elt is a pointer to type Node.

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