CODEWITHSUNDEEP
javaarrays
Reem
Reem

Reputation: 11

Can't understand 3D Arrays

Assume double[][][] x = new double[4][5][6], why x.length, x[2].length, and x[0][0].length are 4, 5, and 6 ? Why [0][0].length is 6?

Upvotes: 1

Views: 920

Answers (4)

aliteralmind
aliteralmind

Reputation: 20163

To visualize a three-dimensional array, and to comprehend its length, you can think of it in terms of literal physical dimensions (remember the book Flat Land?).

One object:

.

An array of objects:

.............

The length of this array (array.length) is how many elements, or dots, counting from left to right.

A two-dimensional array of objects:

.............
.............
.............
.............
.............
.............

Each element in the previous array now has its own array, going down. The length of the first element in this array (array[0].length, the top-left-most) is counting the dots top to bottom.

A three dimensional array is like a cube (this is all assuming uniform lengths among elements)

.............
............. .
............. . .
............. . . .
............. . . .
............. . . .
  . . . . . . . . .
   . . . . . . . ..
    . . . . . . . .

Each element in the previous array now has its own array (going deep "into" the screen). The length of the first element in this array (array[0][0].length, the top-left-most element) is the number of dots deep.

Not once have I (directly) used a three dimensional array.

Upvotes: 5

Sam
Sam

Reputation: 1206

When working with multi-dimensional arrays, it is useful to think of them as an "array-of-arrays." That is to say, double[] is an array of doubles, double[][] would be an array of double[] and double[][][] is an array made of arrays of arrays of doubles. So let's break down your example:

double[][][] x = new double[4][5][6];

Working backwards, we can think of this as being an array of 6 arrays of double[][], each of which contains 5 arrays of double[]. Each of those arrays contains 4 doubles. Confused yet?

Another thing that might help to see the relationship is to write this using parentheses:

(((double[])[])[])

Now, when you access x[2], you are actually retrieving an array of arrays at index 2 of x, and that array has length 5.

When you access x[0][0], this is still just another array (remember, we build up our array as an array of arrays), which you declared in your new statement to have length 6. That is to say, x[0][0] will not contain a number, but will contain array of type double[][], which has a length of 6.

Upvotes: 0

azurefrog
azurefrog

Reputation: 10945

When you declare an array such as double[] x = new double[5] you are saying you want a 5-long array of double, i.e.

    (type) x = new (type)[size];

So when you declare double[][][] x = new double[4][5][6] you are saying you want a 6-long array of double[][]:

    (double[][])[] x = new (double[4][5])[6];

Upvotes: 1

matt forsythe
matt forsythe

Reputation: 3912

With new double[4][5][6] what you have is 20 arrays of length 6. They are addressable as:

x[0][0] // array of length 6
x[0][1] // another array of length 6 here
x[0][2] // etc...
x[0][3]
x[0][4]
x[1][0]
x[1][1]
x[1][2]
x[1][3]
x[1][4]
x[2][0]
x[2][1]
x[2][2]
x[2][3]
x[2][4]
x[3][0]
x[3][1]
x[3][2]
x[3][3]
x[3][4]

Upvotes: 1

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