Reference exercise 2 with call solution

Complete the Primitive types vs references exercises with calls before reviewing the solutions.

Review the primitive exercise solution with AP CS Tutor Brandon Horn.

Original code

public static void main(String[] args)
{
    Coordinate2D e = new Coordinate2D(1,1);
    reference2(e);
    System.out.println(e);
}

private static void reference2(Coordinate2D f)
{
    f = new Coordinate2D(2,2);
}

Output

(1, 1)

Explanation

All arguments in Java are passed by value, including references. The call reference2(e); passes a copy of the value of e to the reference2 method. The value of e is the memory address of the object containing x: 1, y: 1.

The reference2 method changes the actual value of f, which has no effect on the value of e.

Step by step with memory diagrams

Step 1

public static void main(String[] args)
{
    Coordinate2D e = new Coordinate2D(1,1);
    // more code not yet run
}

Memory diagram after Step 1

diagram showing e pointing to a box containing x and y, each with value 1

Except for the variable name, this is the same as Reference exercise 1 with call Step 1.

Step 2

public static void main(String[] args)
{
    Coordinate2D e = new Coordinate2D(1,1);
    reference2(e);
    // more code not yet run
}

private static void reference2(Coordinate2D f)
{
    // more code not yet run
}

Step 2 is immediately after the call to the reference2 method, but before any code inside the reference2 method has been run.

Memory diagram after Step 2

diagram showing e and f pointing to the same box containing x and y, each with value 1

Except for the variable names, this is the same as Reference exercise 1 with call Step 2.

Step 3

public static void main(String[] args)
{
    Coordinate2D e = new Coordinate2D(1,1);
    reference2(e);
    // more code not yet run
}

private static void reference2(Coordinate2D f)
{
    f = new Coordinate2D(2,2);
}

Step 3 is immediately after execution of the statement f = new Coordinate2D(2,2);, but before the reference2 method returns.

Memory diagram after Step 3

diagram showing e pointing to a box containing x with value 1 and y with value 1, and f pointing to a different box containing x with value 2 and y with value 2

The statement f = new Coordinate2D(2,2); creates a new object and points f at it (sets the value of f to the memory address of the object). Changing the actual value of f does not change the value of e. All arguments in Java are passed by value.

Contrast this with the reference1 method in Reference exercise 1 with call. The reference1 method runs a mutator method on the object to which both c and d point.

Step 4

public static void main(String[] args)
{
    Coordinate2D e = new Coordinate2D(1,1);
    reference2(e);
    System.out.println(e);
}

private static void reference2(Coordinate2D f)
{
    f = new Coordinate2D(2,2);
}

Step 4 is after the reference2 method returns and after the print statement executes.

Memory diagram after Step 4

diagram showing e pointing to a box containing x and y, each with value 1

When the reference2 method ends, f no longer exists. In Java, when the last reference to an object is removed, the object is considered garbage and can no longer be accessed.

Output after Step 4

(1, 1)

The main method prints e, which runs the toString method on the object to which e refers.

Additional classes & objects resources

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