c++ creating a static like array with "new" or another way of creating a dynamic array
arrays, c++, new-operator, pointers
Solution
know that the common technique of creating a dynamic array
In C++ that was written 20 years ago, maybe.
These days you should use `std::vector` for dynamic arrays and `std::array` for fixed size array.
If your framework or platform supplies additional array classes (like QT's `QVector`), they are fine too, as long as you don't mess with C-pointers directly, and you have RAII-based array class.
and as for concrete answer, `new T[size]` always returns `T*` , so you cannot catch a pointer returned by `new[]` with `T(*)[size]`.
Problem
I know that the common technique of creating a dynamic array using `new` in C++ is: ``` int * arr = new int[5]; ``` A book also says: ``` short tell[10]; // tell is an array of 20 bytes cout << tell << endl; // displays &tell[0] cout << &tell << endl; // displays address of the whole array short (*p)[10] = &tell; // p points to an array of 20 shorts ``` Now I wonder if there is a way to allocate memory for an array using `new`, so it can be then assigned to a pointer to the whole array. It might look like this: ``` int (*p)[5] = new int[5]; ``` The above example doesn't work. The left side looks correct to me. But I don't know what should be on the right. My intention is to understand if it's possible. And I know that there are `std::vector` and `std::array`. Update: Here is what I actually wanted to check: ``` int (*p1)[5] = (int (*)[5]) new int[5]; // size of the whole array cout << "sizeof(*p1) = " << sizeof(*p1) << endl; int * p2 = new int[5]; // size of the first element cout << "sizeof(*p2) = " << sizeof(*p2) << endl; ``` And here is how to access these arrays: ``` memset(*p1, 0, sizeof(*p1)); cout << "p1[0] = " << (*p1)[0] << endl; memset(p2, 0, sizeof(*p2) * 5); cout << "p2[0] = " << p2[0] << endl; ```