How to rewrite "node->left->key" replacing -> with "(*)." in C++?

Question

I'm new to the "->" symbol, so I'm instead replacing it with (*). . However, when I came across the line of code below, I tried replacing it and it didn't work. What am I doing wrong and is there any way to rewrite it?

I keep getting the error that "key" is a pointer and when I rewrite it, it doesn't work. I have triple checked my code, and yet I still don't understand.

 struct Node{
    int key;
    Node *left;
    Node *right;
 };

  Node* createNode(int key){
   Node *node = new Node();
   (*node).key = key;
   (*node).left = NULL;
   (*node).right = NULL;
   return node;
  }


  int main(){
     Node *root = createNode(1);
     (*root).left = createNode(9);
     cout << root->left->key;  // Correct?
     cout << " OR ";
     cout << ((*root).left).(*key); 
    // this is where my code goes wrong and if I remove the (*) from key
    // and just leave it like .key it's wrong because key has to be a pointer
     return 0;
  }

I expect the output to be "9 OR 9" but it doesn't even let me compile past that point.

Answer 1

If you really want to avoid the -> operator, you can write it like this:

 cout << (*((*root).left)).key;

... but that's painful to write and painful to read, so it makes a great example of why the -> operator is useful :)