Java is Pass by Value, Not Pass by Reference

The following example demonstrates how values are passed in Java.

The example program uses the following class:

public class Balloon {

	private String color;

	public Balloon() {}
	
	public Balloon(String c) {
		this.color = c;
	}
	
	public String getColor() {
		return color;
	}

	public void setColor(String color) {
		this.color = color;
	}
}

The following example program uses a generic method, swap(), that swaps two variables. Another method, changeValue(), attempts to change the variable values.

public class Test {

	public static void main(String[] args) {

		Balloon red = new Balloon("Red"); // memory reference = 50
		Balloon blue = new Balloon("Blue"); // memory reference = 100
		
		swap(red, blue);
		System.out.println("After the swap method executes:");
		System.out.println("`red` color value = " + red.getColor());
		System.out.println("`blue` color value = " + blue.getColor());
		
		changeValue(blue);
		System.out.println("After the changeValue method executes:");
		System.out.println("`blue` color value = " + blue.getColor());
		
	}

	// Generic swap method
	public static void swap(Object o1, Object o2){
		Object temp = o1;
		o1 = o2;
		o2 = temp;
	}

	private static void changeValue(Balloon balloon) { // balloon = 100
		balloon.setColor("Red"); // balloon = 100
		balloon = new Balloon("Green"); // balloon = 200
		balloon.setColor("Blue"); // balloon = 200
	}

}

When you execute the example program, you get the following output:

Output
After the swap method executes:
'red' color value = Red
'blue' color value = Blue
After the changeValue method executes:
'blue' color value = Red

The output shows that the swap() method didn’t swap the color values of the original objects. This helps to show that Java is pass by value, since the swap() method only acts upon copies of the original object reference values.

This swap() method test can be used with any programming language to check whether it’s pass by value or pass by reference.

The Example swap() Method Explained

When you use the new operator to create an instance of a class, the object is created and the variable contains the location in memory where the object is saved.

Balloon red = new Balloon("Red");
Balloon blue = new Balloon("Blue");

Here’s a step-by-step breakdown of what happens when the swap() method executes:

• Assume that red is pointing to memory location 50 and blue is pointing to memory location 100, and that these are the memory locations of both Balloon objects.

• When the class calls the swap() method with the red and blue variables as arguments, two new object variables, o1 and o2, are created. o1 and o2 also point to memory locations 50 and 100 respectively.

• The following code snippet explains what happens within the swap() method:

  public static void swap(Object o1, Object o2) { // o1 = 50, o2 = 100
  	Object temp = o1; // assign the object reference value of o1 to temp: temp = 50, o1 = 50, o2 = 100
  	o1 = o2; // assign the object reference value of o2 to o1: temp = 50, o1 = 100, o2 = 100
  	o2 = temp; // assign the object reference value of temp to o2: temp = 50, o1 = 100, o2 = 50
  } // method terminated
  

• The values of o1 and o2 are swapped, but because the values are copies of the red and blue memory locations, there is no change to the values of the red and blue color values.

Since the variables contain the reference to the objects, it’s a common mistake to assume that you’re passing the reference and Java is pass by reference. However, you’re passing a value which is a copy of the reference and therefore it’s pass by value.

The Example changeValue() Method Explained

The next method in the example program changes the color value of the object referenced by the blue variable:

private static void changeValue(Balloon balloon) { // balloon = 100
	balloon.setColor("Red"); // balloon = 100
	balloon = new Balloon("Green"); // balloon = 200
	balloon.setColor("Blue"); // balloon = 200
}

Here’s a step-by-step breakdown of what happens within the changeValue() method:

• The class calls the changeValue() method on the blue variable that references memory location 100. The first line creates a reference that also points to memory location 100. The color value of the object at memory location 100 is changed to "Red".

• The second line creates a new object (with color value "Green"). The new object is at memory location 200. Any further methods executed on the balloon variable act upon the object at memory location 200, and don’t affect the object at memory location 100. The new balloon variable overwrites the reference created in line 1 and the balloon reference from line 1 is no longer accessible within this method.

• The third line changes the color value of the new Balloon object at memory location 200 to "Blue", but does not affect the original object referenced by blue at memory location 100. This explains why the final line of the example program output prints blue color value = Red, which reflects the change from line 1.

In this article you learned about why Java is pass by value. Continue your learning with more Java tutorials.