Java always passes arguments by value NOT by reference.
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:
-
Declaring a reference named
fof typeFooand assign it to a new object of typeFoowith an attribute"f".Foo f = new Foo("f");
-
From the method side, a reference of type
Foowith a nameais declared and it's initially assigned tonull.public static void changeReference(Foo a)
-
As you call the method
changeReference, the referenceawill be assigned to the object which is passed as an argument.changeReference(f);
-
Declaring a reference named
bof typeFooand assign it to a new object of typeFoowith an attribute"b".Foo b = new Foo("b");
-
a = bis re-assigning the referenceaNOTfto the object whose its attribute is"b".
-
As you call
modifyReference(Foo c)method, a referencecis created and assigned to the object with attribute"f".
-
c.setAttribute("c");will change the attribute of the object that referencecpoints to it, and it's same object that referencefpoints to it.
I hope you understand now how passing objects as arguments works in Java :)