Understanding Array Indexing in Java

Question

I have a brief question about how Java handles arrays. Below is my code:

//import java.util.Arrays;
import static java.lang.System.out;

public class Arrays
{       
    public static void main(String[] args)
    {           
        String [][] multiArray = new String[10][8];

        int k = 1;
        while (k <= 61) {out.print('-'); k++;}
        out.println ();

        for (int i = 0; i < multiArray.length; i++)
        {
            for (int j = 0; j < multiArray[i].length; j++)
            {
                multiArray[i][j] = i + "" + j;
                out.print ("| " + multiArray[i][j] + " ");  
            }
            out.println ("|");
        }           
        k = 1;
        while (k <= 61) {out.print('-'); k++;}
        out.println();          
    }       
}

I understand that you have to create a double "for" loop to print out values for both dimensions and that you have to have:

    multiArray[i].length

so that it knows to reference the length of the second dimension. I just don't understand how it works.

What I'm confused about is this: At the very beginning of the program, directly after I declare my array, if I write a statement like:

    system.out.println (multiArray.length);

It will print the value of 10, which is the length I declared in the first dimension. If I, however, create some random variable like "int a = 0" or "int idontgetthis = 0" and then I write:

    system.out.println (multiArray[a].length);

it somehow knows to print the length of the second dimension, 8. So my question is, how does it know how to do this? It's killing me!! lol

Oliver Charlesworth · Accepted Answer · 2012-07-06 00:15:53Z

5

Because multiArray is really an array of arrays. So multiArray[a] is a reference to an object. That object is itself an array. That array has a length (8), and a property called length which can be used to return that length.

answered Jul 6, 2012 at 0:15

Oliver Charlesworth

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1 Comment

Ryan Christman Over a year ago

An array of arrays? Gaahhhh lol I'm trying my best to understand. Thanks for the hasty response!

jhurtado · Accepted Answer · 2012-07-06 01:00:00Z

0

Basically, it is a concept confusion, by doing:

String[] array;

you are declaring that you will have an array of Strings with an unknown lenght.
A call to: System.out.println(array.length) at this moment will fail with a compilation error because array is not yet initialized (so the compiler can't know how long it is).

By doing:

String[] array = new String[8]

you declare that you will have and array of String and initialize it, specifying it will have space for 8 Strings, the compiler then allocates space for this 8 Strings. Something important to notice is that even when the compiler now knows that you will store 8 Strings in your array, it will fill it with 8 nulls.

So a call to System.out.println(array.length) at this point will return 8 (Compiler knows the size) but a call to System.out.println(array[1]) will return a Null Pointer Exception (You have 8 nulls in it).

Now, in the example you presented, you are declaring a bidimensional array, this is, an array that will contain other arrays.

Bidimensional arrays are initialized as String[][] multiarray = new String[10][8]; and the logic is the same as in simple arrays, the new String[10][8]; indicates the lenght of the array that contains the other arrays, and the new String[10][8]; indicates the length of the contained arrays.

So doing system.out.println(multiArray[x].length); after initializing multiarray is translated as "What is the length of the Xth contained array?", which the compiler, thanks to your initialization, now knows is 8 for all the contained arrays, even when they are full of nulls at the moment.

Hope it helps to add a bit more understanding!

answered Jul 6, 2012 at 1:00

jhurtado

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1 Comment

Ryan Christman Over a year ago

Ah haaaa! Not crystal clear yet, but that is definitely starting to make more sense. I suppose I'm used to thinking about array the way that FORTRAN and C store them...as "x,y" pairings. It's a more mathematical way of thinking about it to me which makes more sense, but oh well. Thank you!

OldCurmudgeon · Accepted Answer · 2012-07-06 01:29:11Z

0

You could try looking at it like this.

public class Arrays{
  public static class EightStrings {
    public String[] strings = new String[8];
  }
  EightStrings[] tenOfThem = new EightStrings[10];
}

answered Jul 6, 2012 at 1:29

OldCurmudgeon

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