Three dim array

[Anonymous]

Hi,
This is a mock exam question:

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:

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,

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

Am I correct here? Can anybody explain this please?
Thanks...

[Jim Yingst]

Look again:
<code><pre> 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]
}
}
};</pre></code>
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.

[This message has been edited by Jim Yingst (edited March 15, 2000).]

[maha anna]

System.out.println( arr[0][1].length ); //Prints NullPointerException - Why??
Since arr[1] must be a ref to another array , and the ref arr[0][1] is set to null means, we have not allocated an array which has to be pointed from this reference arr[0][1]. So attempting to calculate the length of an array which has not yet been allocated memory, because the ref is null here, generates 'NullPointerException'.
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

[This message has been edited by maha anna (edited March 15, 2000).]

[Manju Swamy]

Here is the code to print the array. Any time you see values similar to this
[[[Ljava.lang.String;@f35792c0 (it is not garbage), it is the String representation of the Object
(Arrays are objects in Java). All objects are inherited from Object class. The Object
class has toString() method. When you try to print the object in the
System.out.println() method, it will explicitly call object's toString() method.
The reason for loops are enclosed in the nested try...catch block is to catch the NullPointerException
and print the null values.

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.

[This message has been edited by Manju Swamy (edited March 16, 2000).]

[Anonymous]

Thanks to all for your replies.