Instantiating a 3D array in C++ using parameters

Question

Sorry if this is a noob question, but I'm currently learning C++. I have a function that takes in several parameters - I would like to use those parameters when creating a 3D int array.

void* testFunction(int countX, int countY, int countZ)
{
    const int NX = countX;
    const int NY = countY;
    const int NZ = countZ;

    int* data_out = new int*[NX][NY][NZ]; 
    // The above line throws the error on "NY" - expression must
    // have a constant value
}

From various posts I've learned that you must allocate the array first but I guess I'm doing that wrong? How do you properly initialize a multidimensional array. Also, why does the initialization require a pointer?

Ideally you should use a 1 dimensional vector or unique_ptr<T[]> and fake the three dimensions using math. — NathanOliver
– NathanOliver, Commented Jan 4, 2017 at 19:14
@MichaelO.Your suggestion still throws the same "expression must have a constant value" error. EDIT: The "const int NZ - countZ" was a type - I edited the post to have the proper "=". — Roka545
– Roka545, Commented Jan 4, 2017 at 19:15
Short answer: You can't do this. Long answer: You shouldn't do this. Arrays are the wrong tool, because they do not track their own size. See NathanOliver's comment. — Drew Dormann
– Drew Dormann, Commented Jan 4, 2017 at 19:21
@NathanOliver do any of you know where I can find a simple example that illustrates this concept? — Roka545
– Roka545, Commented Jan 4, 2017 at 19:31

anatolyg · Accepted Answer · 2017-01-04 19:42:49Z

To explain the error: C++ requires a name of a type in its new operator. A name of a type cannot have runtime dimensions, because all types in C++ are static (determined at compilation time).

For example, this allocates 3 elements of type int[4][5]:

new int[3][4][5];

Another example: this allocates NX elements of type int[4][5]:

new int[NX][4][5];

An incorrect example: this would allocate NX elements of type int[NY][NZ], if C++ had support for "dynamic" types:

new int[NX][NY][NZ];

To allocate a 3-dimensional array, or something that looks like it, you can use std::vector:

std::vector<std::vector<std::vector<int>>> my_3D_array;
... // initialization goes here
my_3D_array[2][2][2] = 222; // whatever you want to do with it

To make the syntax less verbose, and streamline the initialization, use typedef (or here using, which is the same):

using int_1D = std::vector<int>;    // this is a 1-dimensional array type
using int_2D = std::vector<int_1D>; // this is a 2-dimensional array type
using int_3D = std::vector<int_2D>; // this is a 3-dimensional array type
int_3D data(NX, int_2D(NY, int_1D(NZ))); // allocate a 3-D array, elements initialized to 0
data[2][2][2] = 222;

If you want to return this array from your function, you should declare it; you cannot just return a void pointer to the data variable. Here is a syntax of a declaration:

using int_1D = std::vector<int>;
using int_2D = std::vector<int_1D>;
using int_3D = std::vector<int_2D>;
int_3D testFunction(int countX, int countY, int countZ)
{
    int_3D data(...);
    ...
    return data;
}

That is, instead of using new, just use std::vector<whatever> as if it were any other type.

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