Pointers. How to create a similar design (in the application)?

Question

C++ / C#

Pointer double **p = 0;
It is necessary to create a construction of the form:

The length of the array is entered previously from the keyboard. Zero out all array elements. Put the number 2 in the first and last elements and print the array to the screen. Delete all dynamic objects.
My code:

double **p = 0;
    int size = 0;
    cout << "Введите размер массива: ";
    size = inputInt();
    cout << "Размер массива: " << size << endl; 
    if (size <= 1) {
        size = 1;
    }
    p = new double*[size];
    for (int i = 0; i < size;i++) {
        p[i] = new double(0);
        cout << p[i] << "\t";
    }
    **p = 2;
    *p[size-1] = 2;
    for (int i = 0; i < size;i++) {
        cout << *p[i] << "\t";
    }    

     for (int i = 0; i < size;i++) {
        delete[] p[i];
    }
    delete[] p;

But I am tormented by vague doubts, I believe that my code is a construction of the form:

Tell me, is it so?
Could you explain to me in words what the 1st construction means?

Answer 1

p -> x -> []

double **p;
p = new double *;
*p = new double[size];
*p[i] = ...

Answer 2

Yes it is. Why do you need a pointer to a pointer in this lab?
How to do the first drawing

p = new (double*)[1];
p[0] = new double[size];

Answer 3

The code

#include <iostream>

using namespace std;

int main()
{
    double **p = 0;

    p = new double *;

    int size;
    cout << "Input size: ";
    cin >> size;

    *p = new double[size];

    for (int i = 0; i < size; i++)
        *(*p + i) = 0;

    **p = *(*p + size - 1) = 2;

    cout << "size: " << size << endl;
    for (int i = 0; i < size; i++)
        cout << *(*p + i) << " ";
    cout << endl;

    delete [] *p;
    delete p;
    p = 0;

    return 0;
}

Conclusion

[[email protected] cpp]$ .iso++ t.cpp -o t
[[email protected] cpp]$ ./t
Input size: 10
size: 10
2 0 0 0 0 0 0 0 0 2 
[[email protected] cpp]$