Java 2D array out of bounds error

Question

I am writing a program that deals with a 2D array, does calculations based on surrounding elements and then creates a new array. To start I created a simple java program to test different cases. I want to correctly deal with if the element is on an edge of the array and if so set the "up, down, left, or right" equal to that element. My test program works for the element being on the top, left, or not on an edge, but not for the bottom or right. This is my current code:

public class ArrayTest 
{ 
   public static void buildE(int[][] array, int y, int x)
   {
      int up; 
      int down; 
      int left;
      int right; 

      //if element is on the top left
      if (y == 0 && x == 0)
      {
         up = array[y][x];
         down = array[y + 1][x];
         left = array[y][x];
         right = array[y][x + 1];
      }

      //if element is on bottom right
      else if (y == array.length && x == array.length)
      {
         up = array[y - 1][x];
         down = array[y][x];
         left = array[y][x - 1];
         right = array[y][x];
      }

      //if element is on top right
      else if(y == 0 && x == array.length)
      {
         up = array[y][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x];
      }

      //if element is on bottom left
      else if (y == array.length && x == 0)
      {
         up = array[y - 1][x];
         down = array[y][x];
         left = array[y][x];
         right = array[y][x + 1];
      }

      //if element is on top 
      else if (y == 0) 
      { 
         up = array[y][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x + 1];
      }  

      //if element is on left
      else if (x == 0)
      {
         up = array[y - 1][x];
         down = array[y + 1][x];
         left = array[y][x];
         right = array[y][x + 1];
      }

      //if element is on bottom
      else if(y == array.length)
      {
         up = array[y - 1][x];
         down = array[y][x];
         left = array[y][x - 1];
         right = array[y][x + 1];
      }

      //if element is on right
      else if (x == array.length)
      {
         up = array[y - 1][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x];
      }

      //if element is not on an edge 
      else
      {
         up = array[y - 1][x];
         down = array[y + 1][x];
         left = array[y][x - 1];
         right = array[y][x + 1];   
      }

      System.out.println();
      System.out.print("#####################################");
      System.out.println();
      System.out.println("Array Element: " + array[y][x]);
      System.out.println("Up: " + up);
      System.out.println("Down: " + down);
      System.out.println("Left: " + left);
      System.out.println("Right: " + right);
   }

   public static void outputArray(int[][] array)
   {
      for(int row = 0; row < array.length; row ++)
      {
         for (int column = 0; column < array[row].length; column++)
            System.out.printf("%d ", array[row][column]);
         System.out.println();
      }
   }

   public static void main(String[] args)
   {
      int [][] myArray = {{1, 12, 13, 14, 15}, {2, 22, 23, 24, 25},
         {3, 32, 33, 34, 35}, {4, 42, 43, 44, 45}, {5, 52, 53, 54, 55}};
      outputArray(myArray);
      buildE(myArray, 4, 0);
   }
}

Furthermore if I set y = 4 it does not recognize this as myArray.length. However I do not think this will be an issue in my actual program if I iterate through until array.length. Any help would be much appreciated!

PermGenError · Accepted Answer · 2012-10-29 22:30:06Z

4

array indexes start from 0, if your array is of length 5, the last elemnt will be accessed at 4th index. in your code

  else if (y == array.length && x == array.length)
{

    up = array[y - 1][x];//Exception here
    down = array[y][x];
    left = array[y][x - 1];
    right = array[y][x];

}

change it to:

 else if (y == array.length-1 && x == array.length-1)
    {

        up = array[y - 1][x];
        down = array[y][x];
        left = array[y][x - 1];
        right = array[y][x];

    }

follow the same approach in other else if's.

edited Oct 29, 2012 at 22:30

answered Oct 29, 2012 at 22:28

PermGenError

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3 Comments

Ryan Sayles Over a year ago

One more question if you don't mind answering, how would I deal with a 2D array that's not a square? I noticed that if it is a rectangle then I'm getting the same error. I believe that is because array.length is not the same for x and y

PermGenError Over a year ago

@rsay3 please post it as a sepeate question

Ryan Sayles Over a year ago

Thank you, I did do that: stackoverflow.com/questions/13177149/… I appreciate all the help!

Andreas Dolk · Accepted Answer · 2012-10-29 22:31:40Z

1

When you call the method with y = 4 and enter

// if element is on left
else if (x == 0) {
    up = array[y - 1][x];
    down = array[y + 1][x];
    left = array[y][x];
    right = array[y][x + 1];
}

then you're actually left and on the bottom line. So you can't go "down". This causes the exception in this case. You could change it to

// if element is on left and not on bottom line
else if (y < array.length && x == 0) {

answered Oct 29, 2012 at 22:31

Andreas Dolk

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