Java always passes arguments by value NOT by reference.

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Let me explain this through an example:

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 to 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 to null.

   public static void changeReference(Foo a)
   

   

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

   changeReference(f);
   

   

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

   Foo b = new Foo("b");
   

   

5. a = b is re-assigning the reference a NOT f to the object whose its attribute is "b".

   

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6. As you call modifyReference(Foo c) method, a reference c is created and assigned to 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 same object that reference f points to it.

   

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