Java variable number of dimensions in an array

Question

Is there a possible way to create an array with a variable number of dimensions?

For example,

int x = 3;
// becomes
int[][][] array = new int[3][3][3];

//and
int y = 4;
//becomes
int[][][][] xray = new int[4][4][4][4];

Part of the reason my examples are so indirect is because I have no idea how one would do something like this.

If I have a variable I would like to create an array with the same number of dimensions as said variable

can you tell why you want to use that, so we can suggest you another approach ? — Daniel, Feb 02 '17 at 02:06
see this http://stackoverflow.com/questions/3104504/is-it-possible-to-dynamically-build-a-multi-dimensional-array-in-java — Jonathan Niu, Feb 02 '17 at 02:06
Can you explain better what do you want to know? The examples are very confusing — nonameable, Feb 02 '17 at 02:07
@nonameable the examples say that if you want 3 dimensions, it will be 3x3x3 array, 4 dimensions, 4x4x4x4 — Daniel, Feb 02 '17 at 02:08
`I have no idea how one would do something like this` ... you have already successfully created multidimensional arrays. What is the actual question? — Tim Biegeleisen, Feb 02 '17 at 02:09
if i have a variable i want to create an array with the number of dimensions as the variable — Mazino, Feb 02 '17 at 02:09
Can you show us the code which you imagine you would use if you had this ability? You might do better to use one of Java's collection classes. — Tim Biegeleisen, Feb 02 '17 at 02:11
So you are looking something like Array q = new Array(x) and you have previously set x as the number of dimensions you want? — nonameable, Feb 02 '17 at 02:14
can you tell us the reason you want this? we certainly can suggest another approach — Daniel, Feb 02 '17 at 02:16
Im trying to solve 2012 CCC Junior 5 "A Coin Game", and I wanted to store data in the most effective way possible. Heres a link to the question http://wcipeg.com/problem/ccc12j5 — Mazino, Feb 02 '17 at 02:18
A friend suggested to store states of coin arrangements numerically, so the arrangement of 4 coins would have a 4 digit number, for example, '1233', the magnitude of each digit corresponding to the position of a coin. The one's place digit refers to the position of coin '1', and the thousands the position of coin '4'. He suggested storing these states in a one-dimensional boolean array(for 4 coins, array length of 3333). With this method, there is still wasted data because array index[1999] can never be accessed(there is no 9th position). — Mazino, Feb 02 '17 at 02:24
This is why I wanted to create a multidimensional array, however there is a variable length of coins, and therefore a variable amount of dimensions in my proposed array — Mazino, Feb 02 '17 at 02:25
you can't do this with Java, but that isn't to say its not a thing in other languages; https://en.wikipedia.org/wiki/Variable-length_array — trooper, Feb 02 '17 at 03:11

score 5 · Answer 1 · answered Feb 02 '17 at 04:22

You can't do this directly, but you can simulate it with a 1-dimensional array.

Suppose you have a 2-dimensional array with 3 rows and 4 columns. You could implement this as an array of 12 elements, and write a get routine to get the A[i,j] element like this:

int[] A = new int[12];

int get(int i, int j) {
    return A[4 * i + j];
}

Of course you could write a set method to set an element of the array in the same way.

Moving on to a 3-dimensional array whose dimensions are 3x4x5, you could do something similar:

int[] A = new int[60];

int get(int i, int j, int k) {
    return A[4*5*i + 5*j + k];
}

or a 4-dimensional 3x4x5x6 array:

int[] A = new int[360];

int get(int i, int j, int k, int m) {
    return A[4*5*6*i + 5*6*j + 6*k + m];
}

And so on... you should be able to see the pattern.

Once you've grasped that, it shouldn't be hard to write a class for an array that takes a variable number of dimensions. The get and set methods could take an int... parameter for a variable number of indexes, and the constructor could similarly take an int... to specify the dimensions. The class would have a private 1-dimensional array whose length is the product of all the dimensions. The get and set methods should check each index to make sure it's >= 0 and less than the corresponding dimension, and throw an exception otherwise.

This is how "true" multi-dimensional arrays are implemented under the hood in most languages that support them (Java is not one of those; it only has 1-dimensional arrays whose elements can be references to other 1-dimensional arrays). It's called "row-major order". See https://en.wikipedia.org/wiki/Row-_and_column-major_order for more information (including the generalized formula).

score 1 · Answer 2 · answered Feb 02 '17 at 02:54

Short answer you can't. Reason: there is no instance in the world of maths where such a construct could be optimum for anything so the Prophets (K&R and all the good ppl who invent code) left such to the developer who needs it.

But here is my 2 cents: I assume you want to initialize your engine...

engine.init(n);

where n is your dimension. Then be able to put and get things from it.

engine.get(a1,a2,a3,...);
engine.put(val,a1,a2,a3...);

Consider a class like this

public class Engine{
private HashMap<Integer[],Object> storage;
private int dimension=1;
private set<integer[]> keys;
public void init(int n){
dimension=n;
//other initialization task;
}

public Object get(int... a){//use var args
integer[] key=findInkeySet(a);
return storage.get(key);
}

public void put(Object val, int... a){
keys.add(a);
storage.put(a,val);
}
}

ofcause there is need for exception handling and all the other generic coding stuf

Deyu瑜 · Answer 3 · 2017-02-02T03:05:47.487

I try to make this , but return type is Object[] ^^"

public Object[] MakeArray(int s){
  ArrayList l = new ArrayList<>(Collections.nCopies(s, 0));
  for (int i = 1; i < s; i++)
    l = copy(l,s);
  return l.toArray();
}
public ArrayList copy (ArrayList l,int s){
    ArrayList<ArrayList> ret = new ArrayList<>(s);
    for (int i = 0; i < s; i++) 
        ret.add(new ArrayList(l));
    return ret;
}

Arrays.toString(MakeArray(2)) = "[[0, 0], [0, 0]]"
Arrays.toString(MakeArray(3)) = "[[[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]]]"

score 0 · Answer 4 · answered Oct 09 '18 at 16:21

public Object createMultidimensionalArray(int dimLength){
    int[] lengths = new int[dimLength];
    for(int i = 0; i < dimLength; i++)
        lengths[i] = dimLength;
    return Array.newInstance(int.class, lengths);
}

Usage example:

int[][][] array3D = (int[][][]) createMultidimensionalArray(3);
System.out.println(Arrays.deepToString(array3D));

Output:

[[[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]], [[0, 0, 0], [0, 0, 0], [0, 0, 0]]]

Explanation:

The function Array.newInstance(Class<?> componentType, int... dimensions) takes as input the wanted array type and lengths of all dimensions.

dimensions is an array that tells the function what size each dimension should be. To get a 4x4x4x4 array, the following code works:

int[] dimensions = {4, 4, 4, 4};
Object arrObj = Array.newInstance(int.class, dimensions);
int[][][][] arr = (int[][][][]) arrObj;

Array.newInstance(...) returns an Object which can be easily converted to the correct type.

Java variable number of dimensions in an array

4 Answers4