Java Pass by Value or Pass by Reference

• Java always passes arguments by value, NOT by reference.

  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 modify 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");
       }
  
  }

  I will explain this in steps:

  1. Declaring a reference named f of type Foo and assign it a new object of type Foo with an attribute "f".

     Foo f = new Foo("f");

     
  2. From the method side, a reference of type Foo with a name a is declared and it's initially assigned null.

     public static void changeReference(Foo a)

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

     changeReference(f);

     
  4. Declaring a reference named b of type Foo and assign it a new object of type Foo with an attribute "b".

     Foo b = new Foo("b");

     
  5. a = b makes a new assignment to the reference a, not f, of the object whose attribute is "b".

     
  6. As you call modifyReference(Foo c) method, a reference c is created and assigned the object with attribute "f".

     
  7. c.setAttribute("c"); will change the attribute of the object that reference c points to it, and it's the same object that reference f points to it.

     

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