problem in dynamic allocation of 3-dimensional array - c++

Question

array_2D = new ushort * [nx];
// Allocate each member of the "main" array
//
for (ii = 0; ii < nx; ii++)
array_2D[ii] = new ushort[ny];
// Allocate "main" array
array_3D = new ushort ** [numexp];
// Allocate each member of the "main" array
for(kk=0;kk<numexp;kk++)
array_3D[kk]= new ushort * [nx];
for(kk=0;kk<numexp;kk++)
for(ii=0;ii<nx;ii++)
array_3D[kk][ii]= new ushort[ny];
the values of numexp,nx and ny is obtained by user..
Is this the correct form for dynamic allocation for a 3d array....We know that the code is working for the 2D array...If this is not correct can anyone suggest a better method?

Answer 1

I think the simplest way to allocate and deal with a multidimensional array is to use one big 1d array (or better yet a std::vector) and provide an interface to index into correctly.
This is easiest to think about first in 2 dimensions. Consider a 2D array with "x" and "y" axis
x=0 1 2
y=0 a b c
1 d e f
2 g h i
We can represent this using a 1-d array, rearranged as follows:
y= 0 0 0 1 1 1 2 2 2
x= 0 1 2 0 1 2 0 1 2
array: a b c d e f g h i
So our 2d array is simply
unsigned int maxX = 0;
unsigned int maxY = 0;
std::cout << "Enter x and y dimensions":
std::cin << maxX << maxY
int array = new int[maxX*maxY];
// write to the location where x = 1, y = 2
int x = 1;
int y = 2;
array[y*maxX/*jump to correct row*/+x/*shift into correct column*/] = 0;
The most important thing is to wrap up the accessing into a neat interface so you only have to figure this out once
(In a similar way we can work with 3-d arrays
z = 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2
y = 0 0 0 1 1 1 2 2 2 0 0 0 1 1 1 0 0 0 1 1 1 2 2 2
x = 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2 0 1 2
array: a b c d e f g h i j k l m n o p q r s t u v w x
Once you figure out how to index into the array correctly and put this code in a common place, you don't have to deal with the nastiness of pointers to arrays of pointers to arrays of pointers. You'll only have to do one delete [] at the end.

Answer 2

Looks fine too me, so long an array of arr[numexp][nx][ny] is what you wanted.
A little tip: you can put the allocation of the third dimension into the loop of the second dimension, aka you allocate each 3rd dimension while the parent subarray gets allocated:
ushort*** array_3D = new ushort**[nx];
for(int i=0; i<nx; ++i){
array_3D[i] = new ushort*[ny];
for(int j=0; j<ny; ++j)
array_3D[i][j] = new ushort[nz];
}
And of course, the general hint: Do that with std::vectors to not have to deal with that nasty (de)allocation stuff. :)
#include <vector>
int main(){
using namespace std;
typedef unsigned short ushort;
typedef vector<ushort> usvec;
vector<vector<usvec> > my3DVector(numexp, vector<usvec>(nx, vector<ushort>(ny)));
// size of -- dimension 1 ^^^^^^ -- dimension 2 ^^ --- dimension 3 ^^
}