Java Pass-by-Value vs. Pass-by-Reference

There has been a good amount of debate on whether “java is pass by value or pass by reference?”. Well, lets conclude it last time, Java is pass by value and not pass by reference. If it had been pass by reference, we should have been able to C like swapping of objects, but we can’t do that in java. We know it already, right?

When you pass an instance to a method, its memory address are copied bit by bit to new reference variable, thus both pointing to same instance. But if you change the reference inside method, original reference will not get change. If it was pass by reference, then it would have got changed also.

To prove it, lets see how memory allocations happen in run time. It should solve the slightest doubts, if any. I am using following program for demonstration of the concept.

public class Foo
{
	private String attribute;

	public Foo (String a){
		this.attribute = a;
	}
	public String getAttribute() {
		return attribute;
	}
	public void setAttribute(String attribute) {
		this.attribute = attribute;
	}
}

public class Main
{
     public static void main(String[] args){
          Foo f = new Foo("f");
          changeReference(f); // It won't change the reference!
          modifyReference(f); // It will change the object that the reference variable "f" refers to!
     }
     public static void changeReference(Foo a) {
          Foo b = new Foo("b");
          a = b;
     }
     public static void modifyReference(Foo c) {
          c.setAttribute("c");
     }
}

Lets see what happen on runtime step by step :

1) Foo f = new Foo(“f”);

This statement will create an instance of class Foo, with ‘attribute’ initialized to ‘f’. The reference to this created instance is assigned to variable f;

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2) public static void changeReference(Foo a)

When this executes then a reference of type Foo with a name a is declared and it’s initially assigned to null.

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3) changeReference(f);

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

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4) Foo b = new Foo(“b”); inside first method

This will do exactly the same as in first step, and will create a new instance of Foo, and assign it to b;

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5) a = b;

This is the important point. Here, we have three reference variables and when statement executes, a and b will point to same instance created inside the method. Note: f is unchanged and it is continually pointing to instance, it was pointing originally. NO CHANGE !!

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6) modifyReference(Foo c);

Now when this statement executed a reference, c is created and assigned to the object with attribute “f”.

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7) c.setAttribute(“c”);

This will change the attribute of the object that reference c points to it, and its same object that reference f points to it.

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I hope that this explanation was enough clear to make your understanding better, if it was not already.

Happy Learning !!

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13 thoughts on “Java Pass-by-Value vs. Pass-by-Reference”

  1. In step 5, the value is actually changed within the scope of the changeReference() function.
    If you were to print out the value inside the scope of the function, the value will be “b”.
    This is because the variable a only has local scope since it is declared in the signature of the function. At the end of the function, a goes out of scope, unhiding/unshadowing the original variable.

    While this example is correct, I think the confusion with Java isn’t PassByValue vs. PassByReference. Its PassByReference vs. Using Objects By Reference. Java doesn’t pass values as arguments by reference, however it does very much .use objects by reference for performance optimization reasons.

    Reply
    • This is correct. Here is what happens:

      Original memory address of f: Foo@38cccef
      Original value of f: f

      Inside changeRef scope, before change:
      Memory address = Foo@38cccef
      Value = f

      Inside changeRef scope, after change:
      Memory address = Foo@64b8f8f4
      Value = b

      Inside modifyRef scope, before change:
      Memory address = Foo@38cccef
      Value = f

      Inside modifyRef scope, after change:
      Memory address = Foo@38cccef
      Value = c

      Reply
  2. Very good explanation

    I have confusion in the point 2 line “initially assigned to null”

    But when i print the reference before assignment initially it point the reference of f.

    public static void changeReference(PassByValue a) {
    		System.out.println("Before Reference of a "+a);
    		PassByValue b = new PassByValue("b");
    		a = b;
    		System.out.println("Reference of a "+a);
    		System.out.println("Reference of b "+b);
    	}
    

    Please correct me if i am wrong.

    Reply
    • It will not be assigned to NULL. Actually a reference is created and the instance is assigned to it and passed to the method.

      Foo a = f; — this will happen behind and passed the reference to the method

      Reply

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