Shallow copy process of C++ copy constructor?

//

#include "stdafx.h"
#include <iostream>
using namespace std;

class MyClass
{
public:
    MyClass()
    {
        pnumber = new int(5);
    }

    ~MyClass()
    {
        delete pnumber;
    }

    int number = 0;
    int *pnumber = nullptr;
};

int main()
{

    MyClass a;
    a.number = 300;
    cout << "value of a.n" << a.number << endl;
    cout << "a.Value of pn" << *a.pnumber << endl;
    cout << "a.Address value of pn" <<a.pnumber << endl;

    MyClass b(a);
    cout << "b.The value of n" <<b.number << endl;
    cout << "b.The value of pn" << *b.pnumber << endl;
    cout << "b.Address value of pn" <<b.pnumber << endl;

    a.number = 400;

    return 0;
}

I'm still an undergraduate, so I don't know exactly how to ask questions, so I apologize for the difficulty with the title and questions.

In the above code, instance b copies a through the default copy constructor, which I learned is

Because the default copy generator is a shallow copy machine, b as shown in the figure.The pnumber variable is a.It is copied in the form of pointing to new int (5) indicated by pnumber.

At this time, since the destructor contains the delete number, it is like releasing the already released memory (red arrow) again, resulting in an error.

That's what I know.

If you're curious about this,

a.pnumber is explicitly assigned by the constructor's pnumber = new int(5) code, b. generated by the copy constructor.Does number really get assigned (before it becomes a copy generator as shown)? If so, how and when will you receive it? It's about.

When an error occurs

(1) b.pnumber is assigned new int(5) (red arrow) -> Memory (red arrow) release due to shallow copy -> Destructor code requests release again for already released area, resulting in error

(2) b.The pnumber is a upon creation.Shallow copy of pnumber -> Never newly assigned -> Destructor code requests release of unallocated area in the first place, resulting in error

I wonder which of these two processes is correct.

In books, lectures, etc., it is already at some point b before shallow copying, as shown in (1).I asked you because there is no explanation about the premise that the pnumber was assigned.

Thank you.