Is Java Pass by Reference or Pass by Value?

 
Java manipulates objects ‘by reference’, but it passes object references to methods ‘by value’.
 
Consider the following program:
 

public class MainJava
{
	public static void main(String[] arg) {
		Foo f = new Foo("f");
		// It won't change the reference!
		changeReference(f);
		// It will modify the object that the reference variable "f" refers to!
		modifyReference(f);
	}

	public static void changeReference(Foo a) {
		Foo b = new Foo("b");
		a = b;
	}

	public static void modifyReference(Foo c) {
		c.setAttribute("c");
	}
}

 
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I will explain this in steps:

1. Declaring a reference named fof type Fooand assign it to a new object of type Foowith an attribute "f".
    

   Foo f = new Foo("f");

    

   

2. From the method side, a reference of type Foowith a name ais declared and it’s initially assigned to null.
    

   public static void changeReference(Foo a)

    
   

3. As you call the method changeReference, the reference awill be assigned to the object which is passed as an argument.

   changeReference(f);

    
   

4. Declaring a reference named bof type Fooand assign it to a new object of type Foowith an attribute "b".

   Foo b = new Foo("b");

    
   

5. a = b is re-assigning the reference aNOT fto the object whose its attribute is "b".
   That is Only the references of a is modified, not the original ones(Reference of fremains the same)

   

6. As you call modifyReference(Foo c)method, a reference cis created and assigned to the object with attribute "f".
    
   
7. c.setAttribute(“c”); will change the attribute of the object that reference cpoints to it, and it’s same object that reference fpoints to it.
    
   

I hope you understand now how passing objects as arguments works in Java :)

About Mohaideen Jamil