Why the last value appear in hexadecimal rather than decimal? c++

Question

struct Node

{

Node* next;

int num;

};

int main()

{

Node* ptr = new Node;

ptr->num = 11;

ptr->next = NULL;

cout << ptr << "\n";

cout << &ptr->next << "\n";

cout << "---------------------\n";

cout << &ptr->num << "\n";

cout << &ptr->next + 1 << "\n";

cout << "---------------------\n";

cout << *(&ptr->num) << "\n";

cout << *(&ptr->next + 1) << "\n";

}

00B1E358

00B1E358

---------------------

00B1E35C

00B1E35C

---------------------

11

0000000B

Answer 1

Hello, Hussain K.

The answer to your question is "because of types". Please allow me to explain. The last 2 statements are:

cout << *(&ptr->num) << "\n";

cout << *(&ptr->next + 1) << "\n";

The first of these does the following:

1. Take the address of the int reached by the expression `ptr->num`.
2. Dereference the address of the int, returning an int.
3. Send the int to cout.

The second of these does something similar, but different:

1. Take the address of the Node * reached by the expression `ptr->next`.
2. Add 1 to the address of the Node *, returning the address of another Node *(!)
3. Dereference the address of the Node *, returning a Node *(!)
4. Send the Node * to cout.

So, the first is streaming the value of an int (which defaults to decimal), and the second is streaming the value of a (Node *) pointer (which defaults to hexadecimal). Even though the addresses being dereferenced are the *same*, the *types* are different, as is the behavior of the stream conversion of int/pointer to characters.

I hope this clears it up for you. Please comment if you're not comfortable with my explanation.

Answer 2

What has higher priority, the address operator & or the dereference operator -->