Question

I am a c++ newbie. While learning I came across this.

if I have a pointer like this

int (*a)[2][3]

cdecl.org describe this as declare a as pointer to array 2 of array 3 of int:

When I try

int x[2][3];

a = &x;

this works.

My question is how I can initialize a when using with new() say something like

a = new int [] [];

I tried some combinations but doesn’t get it quite right.

Any help will be appreciated.

Answer 1

It’s probably not the answer you’re looking for, but what you
need is a new expression whose return type is (*)[2][3] This
is fairly simple to do; that’s the return type of new int
[n][2][3], for example. Do this, and a will point to the
first element of an array of [2] of array of [3] int. A three
dimensional array, in sum.

The problem is that new doesn’t return a pointer to the top
level array type; it returns a pointer to the first element of
the array. So if you do new int[2][3], the expression
allocates an array of 2 array of 3 int, but it returns
a pointer to an array of 3 int (int (*a)[3]), because in C++,
arrays are broken (for reasons of C compatibility). And there’s
no way of forcing it to do otherwise. So if you want it to
return a pointer to a two dimensional array, you have to
allocate a three dimensional array. (The first dimension can be
1, so new [1][2][3] would do the trick, and effectively only
allocate a single [2][3].)

A better solution might be to wrap the array in a struct:

struct Array
{
    int data[2][3];
};

You can then use new Array, and everything works as expected.
Except that the syntax needed to access the array will be
different.