linked lists and pointers clarify

[4ET0pfjN]

let's say I'm building a BST (binary search tree) w/ node as such:

struct bst_node//Binary search tree node
{
 int data;
 bst_node* left;
 bst_node* right;
};

int main()
{
 bst_node* root = NULL;
 
 return 0;
}

So for outputs for point 3, I sort of understand why it doesn't work when we use dereference operator...:
1) if it was: cout << root << endl;//This would be address of what pointer root pts to which is address 0

2) if it was: [code]cout << &root << endl;//This would be the address of the the declared root pointer itself since a pointer is a variable and it takes up space so it must exist somewhere in memory

3) if it was: [code] cout << *root << endl;//program crashes, is this b/c a root is not exactly a pointer...I'm not sure how to explain so that's why we have to use right arrow operator to access members: data, left, right.

[vlad from moscow]

This statement

cout << *root << endl;

shall not be compiled because the operator << was not defined for the type struct bst_node.

If to accept that this operator was defined then *root shall be valid object that is it shall be defined in the memory.

[4ET0pfjN]

Can you clarify a little more

[vlad from moscow]

What do you want to know?

[4ET0pfjN]

I thought it was b/c of the fact that a bt_node is a special pointer for the way I'm using it but I guess I shouldn't go too deep into details unless it's useful as a programmer, I don't want to get into history I mean...so I'll leave it at that,thanks