Complete the Arrays as objects exercises before reviewing the solutions.
Review the Arrays as objects exercise 4 solution with AP CS Tutor Brandon Horn.
Original code
public static void main(String[] args)
{
Coordinate2D[] coors = new Coordinate2D[3];
coors[0] = new Coordinate2D(1, 1);
coors[1] = new Coordinate2D(2, 2);
coors[2] = new Coordinate2D(3, 3);
System.out.println(Arrays.toString(coors));
// prints: [(1, 1), (2, 2), (3, 3)]
mystery4(coors);
System.out.println(Arrays.toString(coors));
}
public static void mystery4(Coordinate2D[] coorsInside)
{
coorsInside = new Coordinate2D[3];
coorsInside[0] = new Coordinate2D(5, 5);
}
Output
[(1, 1), (2, 2), (3, 3)]
[(1, 1), (2, 2), (3, 3)]
Explanation
The mechanics are identical to those in Exercise 2.
The statement coorsInside = new Coordinate2D[3]; changes the actual value of coorsInside, which has no effect on the value of coors.
Step by step memory diagram
Step 1
public static void main(String[] args)
{
Coordinate2D[] coors = new Coordinate2D[3];
coors[0] = new Coordinate2D(1, 1);
coors[1] = new Coordinate2D(2, 2);
coors[2] = new Coordinate2D(3, 3);
System.out.println(Arrays.toString(coors));
// prints: [(1, 1), (2, 2), (3, 3)]
// additional code not yet run
}
Memory diagram after Step 1
This is the same as in the Exercise 3 solution.
Output after Step 1
[(1, 1), (2, 2), (3, 3)]
Step 2
public static void main(String[] args)
{
Coordinate2D[] coors = new Coordinate2D[3];
coors[0] = new Coordinate2D(1, 1);
coors[1] = new Coordinate2D(2, 2);
coors[2] = new Coordinate2D(3, 3);
System.out.println(Arrays.toString(coors));
// prints: [(1, 1), (2, 2), (3, 3)]
mystery4(coors);
// additional code not yet run
}
public static void mystery4(Coordinate2D[] coorsInside)
{
// additional code not yet run
}
Memory diagram after Step 2
This is also the same as in the Exercise 3 solution.
Output after Step 2
[(1, 1), (2, 2), (3, 3)]
Step 3
public static void main(String[] args)
{
Coordinate2D[] coors = new Coordinate2D[3];
coors[0] = new Coordinate2D(1, 1);
coors[1] = new Coordinate2D(2, 2);
coors[2] = new Coordinate2D(3, 3);
System.out.println(Arrays.toString(coors));
// prints: [(1, 1), (2, 2), (3, 3)]
mystery4(coors);
// additional code not yet run
}
public static void mystery4(Coordinate2D[] coorsInside)
{
coorsInside = new Coordinate2D[3];
// additional code not yet run
}
Memory diagram after Step 3
As in Exercise 2, the statement ... = new Coordinate2D[3]; creates a new array (in this case containing the values [null, null, null]). The statement coorsInside = ... changes the actual value of coorsInside. Changing the value of coorsInside does not effect the value of coors. coors still points to the original array, which remains unchanged.
Output after Step 3
[(1, 1), (2, 2), (3, 3)]
Step 4
public static void main(String[] args)
{
Coordinate2D[] coors = new Coordinate2D[3];
coors[0] = new Coordinate2D(1, 1);
coors[1] = new Coordinate2D(2, 2);
coors[2] = new Coordinate2D(3, 3);
System.out.println(Arrays.toString(coors));
// prints: [(1, 1), (2, 2), (3, 3)]
mystery4(coors);
// additional code not yet run
}
public static void mystery4(Coordinate2D[] coorsInside)
{
coorsInside = new Coordinate2D[3];
coorsInside[0] = new Coordinate2D(5, 5);
}
Memory diagram after Step 4
As in Exercise 2, the statement coorsInside[0] = ... changes the value at index 0 in the array to which coorsInside points. This has no effect on coors, the array to which coors points, or any of the objects inside the array to which coors points.
Output after Step 4
[(1, 1), (2, 2), (3, 3)]
Step 5
public static void main(String[] args)
{
Coordinate2D[] coors = new Coordinate2D[3];
coors[0] = new Coordinate2D(1, 1);
coors[1] = new Coordinate2D(2, 2);
coors[2] = new Coordinate2D(3, 3);
System.out.println(Arrays.toString(coors));
// prints: [(1, 1), (2, 2), (3, 3)]
mystery4(coors);
System.out.println(Arrays.toString(coors));
}
Memory diagram after Step 5
When mystery4 returns (end), coorsInside (the parameter) goes away. Since nothing points ot the array containing [(5, 5), null, null], it goes away too. coors points to the original array, that still contains the original values.
Output after Step 5
[(1, 1), (2, 2), (3, 3)]
[(1, 1), (2, 2), (3, 3)]