posted March 15, 2000 09:36 PM         

code:

---

public class ThreeDimArray
{
    public static void main(String args[])
    {
        String[][][] arr =
        {
            { {}, null },
            { { "1", "2" }, { "1", null, "3" } },
            {},
            { { "1", null } }
        };
        System.out.println( arr.length + arr[1][2].length ); 
    }
}

---

Select valid answer:
a. ArrayIndexOutOfBoundsException //ans
b. NullPointerException
c. 4
d. 7

I tried the code and it gives ArrayIndexOutOfBoundsException

I have modified the code to display all the values in the array:

code:

---

public class ThreeDimArray
{
    public static void main(String args[])
    {
        String[][][] arr =
        {
            { {}, null },
            { { "1", "2" }, { "1", null, "3" } },
            {},
            { { "1", null } }
        };
      
     //  System.out.println(arr.length + arr[1][2].length ); // Prints ArrayIndexOutOfBoundsException
      
      	System.out.println(arr.length); // Prints 4
      	
        System.out.println( arr[0][0].length ); // Prints 0
     //  System.out.println( arr[0][0][0] ); //Prints ArrayIndexOutOfBoundsException
        System.out.println( arr[0][0] ); // prints junk values
        System.out.println( arr[0] ); // prints junk values
     //  System.out.println( arr[0][1].length ); //Prints NullPointerException - Why?? IMO - length should be 0 
      	System.out.println( arr[0][1] ); // prints null
        System.out.println( arr[1][0].length ); // Prints 2
        System.out.println( arr[1][0][0] ); // Prints 1
        System.out.println( arr[1][0][1] ); // Prints 2
        System.out.println( arr[1][1].length ); // Prints 3
		System.out.println( arr[1][1][0] ); // Prints 1
        System.out.println( arr[1][1][1] ); // Prints null
        System.out.println( arr[1][1][2] ); // Prints 3
        System.out.println( arr[2].length ); // Prints ArrayIndexOutOfBoundsException - Why? IMO - length should be 0
        System.out.println( arr[3][0].length ); // Prints 2
        System.out.println( arr[3][0][0] ); // Prints 1
        System.out.println( arr[3][0][1] ); // Prints null
    }
}

---

The question I have is with this line of code
System.out.println( arr[0][1].length ); //Prints NullPointerException - Why?? IMO - length should be 1, since arr[0][1][0] contains null value.
Also,
in the line: System.out.println( arr[2].length ); // Prints ArrayIndexOutOfBoundsException - Why? IMO - length should be 0, because,

code:

---

String arr1 = {};
System.out.println( arr.length ); // Prints length as 0

---

Am I correct here? Can anybody explain this please?

Thanks...

posted March 15, 2000 11:03 PM            

    String[][][] arr =
    {   // arr[0]
        {   // arr[0]
            {}, // arr[0][0]
            null // arr[0][1]
        },     
        {   // arr[1]
            {   // arr[1][0]
                "1",  // arr[1][0][0]
                "2"   // arr[1][1][1]
            }, 
            {   // arr[1][1]
                "1",  // arr[1][1][0]
                null, // arr[1][1][1]
                "3"   // arr[1][1][2]
            } 
        }, 
        {}, // arr[2]                                  
        {   // arr[3]
            {   // arr[3][0]
                "1",   // arr[3][0][0]
                 null  // arr[3][0][1]
            } 
        }                     
    };

The red null corresponds to arr[0][1]. At this point in the array, the third dimension does not exist, because there's no array there to hold it in. Note that there are no extra parentheses around the null which would justify indenting it one more level.

As for your second question: when I run it, the line prints 0 as expected. I think you're mistaken.

posted March 15, 2000 11:18 PM            

System.out.println( arr[2].length ); // Prints ArrayIndexOutOfBoundsException - Why?
No length of arr[2] is zero . 0 is printed. No exception will be thrown. Please check again.
Since here arr[2] must be an array of arrays, and this code sets arr[2] to just an empty array, which does not have any arrays inside , trying to calculate the length of an empty array prints 0.

regds
maha anna

posted March 16, 2000 11:58 AM         

The reason for loops are enclosed in the nested try...catch block is to catch the NullPointerException
and print the null values.

code:

---

public class ThreeDimArray {
    public static void printArray(String[][][] myArray) {
        int i = 0, j = 0, k =0;
        try {
            System.out.println("length:" + myArray.length + " value:" + myArray);
            for (i = 0; i < myArray.length; i++) {
                System.out.println("i:" + i + " length[i]:" + myArray[i].length + " value[i]:" + myArray[i]);
                if (myArray[i].length == 0) continue;
                try {
                    for (j = 0; j < myArray[i].length; j++) {
                        System.out.println("i:" + i + " j:" + j + " length[i][j]:" + myArray[i][j].length + " value[i][j]:" + myArray[i][j]);
                        if (myArray[i][j].length == 0) continue;
                        try {
                            for (k = 0; k < myArray[i][j].length; k++) {
                                System.out.println("i:" + i + " j:" + j + " k:" + k + " value[i][j][k]:" + myArray[i][j][k]);
                                if (myArray[i][j].length == 0) continue;
                                System.out.println("Array [" + i + "][" + j + "][" + k + "] = " + myArray[i][j][k]);
                            }
                        } catch (NullPointerException npe) {
                            System.out.println("Array [" + i + "][" + j + "][" + k + "] = " + null);
                            continue;
                        }
                    }
                } catch (NullPointerException npe) {
                    System.out.println("Array [" + i + "][" + j + "] = " + null);
                    continue;
                }
            }
        } catch (NullPointerException npe) {
            System.out.println("Array [" + i + "] = " + null);
        }
    }
    public static void main(String args[])    {
        String[][][] arr =  {
            { {}, null },
            { { "1", "2" }, { "1", null, "3" } },
            {},
            { { "1", null } }
        };
        printArray(arr);
    }
}

---

Go thorough the program output and analyze. Let us know if you need any further clarifications.