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
f
of typeFoo
and assign it to a new object of typeFoo
with an attribute"f"
.Foo f = new Foo("f");
-
From the method side, a reference of type
Foo
with a namea
is declared and it's initially assigned tonull
.public static void changeReference(Foo a)
-
As you call the method
changeReference
, the referencea
will be assigned to the object which is passed as an argument.changeReference(f);
-
Declaring a reference named
b
of typeFoo
and assign it to a new object of typeFoo
with an attribute"b"
.Foo b = new Foo("b");
-
a = b
is re-assigning the referencea
NOTf
to the object whose its attribute is"b"
.
-
As you call
modifyReference(Foo c)
method, a referencec
is created and assigned to the object with attribute"f"
. -
c.setAttribute("c");
will change the attribute of the object that referencec
points to it, and it's same object that referencef
points to it.
I hope you understand now how passing objects as arguments works in Java :)