Complete the Primitive types vs references exercises before reviewing the solutions.
Review the reference exercise 2 solution with AP CS Tutor Brandon Horn.
Original code
Coordinate2D referenceOne = new Coordinate2D(1, 1);
Coordinate2D referenceTwo = referenceOne;
System.out.println(referenceOne);
System.out.println(referenceTwo);
Output
(1, 1)
(1, 1)
Step by step with memory diagrams
Step 1
Coordinate2D referenceOne = new Coordinate2D(1, 1);
Memory diagram after Step 1
This is the same first line as in Reference exercise 1, so the memory diagram is the same.
Step 2
Coordinate2D referenceOne = new Coordinate2D(1, 1);
Coordinate2D referenceTwo = referenceOne;
Memory diagram after Step 2
On the 2nd line, the right side is evaluated first. The right side is referenceOne, which means the value of referenceOne. The value of referenceOne is the memory address of the object containing x: 1, y: 1. In the diagram, the value of referenceOne is the arrow.
The left side, Coordinate2D referenceTwo, makes a new variable/reference. The value of referenceTwo is set to a copy of the value of referenceOne.
referenceOne and referenceTwo point to the same object containing x: 1, y: 1.
Step 3
Coordinate2D referenceOne = new Coordinate2D(1, 1);
Coordinate2D referenceTwo = referenceOne;
System.out.println(referenceOne);
System.out.println(referenceTwo);
Memory diagram after Step 3
Output after Step 3
(1, 1)
(1, 1)
Both print statements print the same object, so the same output is produced twice. See Reference exercise 1 Step 3 for an explanation of what happens when referenceOne is printed.