What is the type of a pointer to a 2D array?

Question

I know that the following is not correct:

int arr[2][3] = {}; //some array initialization here int** ptr; ptr = arr; 

But I am quite surprised that the following lines actually work

int arr[2][3] = {}; //some array initialization here auto ptr = arr; int another_arr[2][3] = {}; //some array initialization here ptr = another_arr; 

Can anyone possibly explain what is the type assigned to ptr in the second block of code, and what happened underneath?

Answer 1

Well, arrays decay to pointers when used practically everywhere. So naturally there's decay going on in your code snippet too.

But it's only the "outer-most" array dimension that decays to a pointer. Since arrays are row-major, you end up with int (*)[3] as the pointer type, which is a pointer to a one-dimensional array, not a two dimensional array. It points to the first "row".

If you want ptr's deduction to be a pointer to the array instead, then use the address-of operator:

auto ptr = &arr; 

Now ptr is int(*)[2][3].

Answer 2

In

auto ptr = arr; 

arr decays into a pointer to its first element in the normal way; it's equivalent to

auto ptr = &arr[0]; 

Since arr[0] is an array of three ints, that makes ptr a int (*)[3] - a pointer to int[3].

another_arr decays in exactly the same way, so in

ptr = another_arr; 

both sides of the assignment have the type int (*)[3], and you can assign a T* to a T* for any type T.

A pointer to arr itself has type int(*)[2][3].

If you want a pointer to the array rather than a pointer to the array's first element, you need to use &:

auto ptr = &arr;