Java 2D arrays, understanding output

Question

I got this question from a previous exam paper. I am trying to understand how do you arrive to the answer: 518

Consider the following Java program, consisting of the class Pins. Notice this is defined to have one constructor and one method named run. Both have no parameters.

class Pins {
    private int[][] ints = {{0,1,2}, {3,4,5},{6,7,8}};

    public Pins()
    {
        int[] a = ints[0];
        ints[0][2] = ints[0][1];
        ints[0] = ints[1];
        ints[1] = a;
    }

    public void run()
    {
        int i = -1;
        while (++i < ints.length)
        {
            int total = 1;
            for (int j = 0; j < ints.length; j++) {
                if (j % 2 == 0) {
                    total += ints[i][j];
                } else {
                    total -= ints[i][j];
                }
            }

            System.out.println(total);
        }
    }
}

Given an instance of this class, what is printed out by its run method? Use diagrams to justify and explain your answer.

>Why does a become {0,1,1} after the second step in the constructor and not {3,4,5} in the third step in the constructor?

Answer 1

The constructor executes like so:

int[] a = ints[0];       // ints: {a: {0,1,2}, {3,4,5}, {6,7,8}}
ints[0][2] = ints[0][1]; // ints: {a: {0,1,1}, {3,4,5}, {6,7,8}}
ints[0] = ints[1];       // ints: {{3,4,5}, {3,4,5}, {6,7,8}}; a: {0,1,1}
                         //         \-same-array/                         
ints[1] = a;             // ints: {{3,4,5}, {0,1,1}, {6,7,8}}                  

And then in run() , what it does, is, for every row, it computes:

1 + row[0] - row[1] + row[2];

So:

1 + 3 - 4 + 5 = 5
1 + 0 - 1 + 1 = 1
1 + 6 - 7 + 8 = 8

What's potentially tricky is that ints contains references to the rows, so when you say a = ints[0] , a points to the row, rather than containing a copy of it.

Answer 2

You see it better, if you write down the single results (just some System.out.println(...); added):

class Pins
{
private int[][] ints = {{0,1,2}, {3,4,5},{6,7,8}};
public Pins()
   {
    int[] a = ints[0];
    ints[0][2] = ints[0][1];
    ints[0] = ints[1];
    ints[1] = a;
   }
public void run()
   {
       for(int i=0; i < ints.length; i++){
           for(int j=0; j < ints[i].length;j++){
               System.out.print(ints[i][j]+";");
           }
           System.out.println("");
       }
   int i = -1;
   while (++i < ints.length)
      {
       int total = 1;
       for (int j = 0; j < ints.length; j++)
          {
          if (j % 2 == 0)
             {
              total += ints[i][j];
             } else
          {
           total -= ints[i][j];
          }
           System.out.println("Total:"+total);
       }
       System.out.println(total);
    }
  }

public static void main(String[] args){
    Pins pin = new Pins();
    pin.run();
}
}

results:

3;4;5;    //ints[0] after swap
0;1;1;    //ints[1] after swap
6;7;8;    //ints[2] after swap
Total:4
Total:0
Total:5
5
Total:1
Total:0
Total:1
1
Total:7
Total:0
Total:8
8

Answer 3

Constructor changes ints[][] around:

int[] a = {0,1,2} //references ints[0]
ints[0][2] = ints[0][1] // ints[][] = {{0,1,1},{3,4,5},{6,7,8}}
// a = {0,1,1} since it references ints[0]
ints[0] = ints[1] // ints[][] = {{3,4,5},{3,4,5},{6,7,8}} 
//changes reference of ints[0], a[] still pointing to old reference
ints[1] = a // ints[][] = {{3,4,5},{0,1,1},{6,7,8}
//points ints[1] back to a's reference
//final ints: {{3,4,5},{0,1,1},{6,7,8}

i starts as -1.
++i increments before comparing, so the first while check is 0 < ints.length (which is currently 3). i is now 0 total is 1

for j = 0; j < 3; j++
j = 0
    if (j%2 == 0) //true
        total += 3 //total = 4
j = 1
    if (j%2 == 0) //false
        total -= ints[0][1] //4, total = 0
j = 2
    if (j%2 == 0) //true
        total += ints[0][2] //5, total = 5
j = 3
    Print "5"

i is now 1 total is 1

for j = 0; j < 3; j++
j = 0
    if (j%2 == 0) //true
        total += ints[1][0] //0, total = 1
j = 1
    if (j%2 == 0) //false
        total -= ints[1][1] //1, total = 0
j = 2
    if (j%2 == 0) //true
        total += ints[1][2] //1, total = 1
j = 3
    Print "1"

i is now 2 total is 1

for j = 0; j < 3; j++
j = 0
    if (j%2 == 0) //true
        total += ints[2][0] //6, total = 7
j = 1
    if (j%2 == 0) //false
        total -= ints[2][1] //7, total = 0
j = 2
    if (j%2 == 0) //true
        total += ints[2][2] //8, total = 8
j = 3
    Print "8"

Final output:

5

1

8

(Println will print on separate lines, so for "518" you'd need to just System.out.print())

Answer 4

after getting instance of pin this array look like:

{3,4,5}

{0,1,1}

{6,7,8}

when i = 0(traversing row 0) total = 1+3-4+5=5

when i = 1(traversing row 1) total = 1+0-1+1=1

when i = 2(traversing row 2) total = 1+6-7+8=8