It is not passed by reference. Rather, it is passed by value of the reference, which is a subtle but important distinction.
After you mutate foo
in the rest of your main()
method, the foo
field will also exhibit these mutations, as you state, since both variables point to the same instance. However, if you reassign foo
to something new, the foo
field will not be changed. This would not be true if foo
were truly passed by reference. In short, everything is passed by value in Java; it just so happens that objects are dealt with by reference, and so the values of these references are passed.
I'll try to illustrate this with an example. Consider the following class:
class A {
public int n;
public A(int n) {
this.n = n;
}
}
and the following method:
public static void mutate(A a) {
a.n = 42;
}
Now we can have something like this:
A a = new A(0);
A.mutate(a);
System.out.println(a.n);
42
We can see that the state of a
was changed in mutate()
. Now let's modify the static method:
public static void mutate(A a) {
a = new A(42);
}
and try again:
A a = new A(0);
A.mutate(a);
System.out.println(a.n);
0
As you can see, the state of a
is unchanged. If the reference had been passed to the function, we would expect the effect of the reassignment to be evident beyond the scope of the method. Nevertheless, some reference was in fact passed, since mutating the argument caused changes outside of the method as well.