From James Gosling in "The Java Programming Language":
"...There is exactly one parameter passing mode in Java - pass by value - and that keeps things simple. .."
Java is always pass by value.
When you pass a primitive it's a copy of the value, when you pass an object it's a copy of the reference pointer.
I don't think you can. Your best option might be to encapsulate the thing you want to pass "by ref" onto another class instance, and pass the (outer) class's reference (by value). If you see what I mean...
i.e. your method changes the internal state of the object it is passed, which is then visible to the caller.
Another option is to use an array, e.g.
void method(SomeClass[] v) { v[0] = ...; }
but 1) the array must be initialized before method invoked, 2) still one cannot implement e.g. swap method in this way...
This way is used in JDK, e.g. in java.util.concurrent.atomic.AtomicMarkableReference.get(boolean[])
.
In Java there is nothing at language level similar to ref. In Java there is only passing by value semantic
For the sake of curiosity you can implement a ref-like semantic in Java simply wrapping your objects in a mutable class:
public class Ref<T> {
private T value;
public Ref(T value) {
this.value = value;
}
public T get() {
return value;
}
public void set(T anotherValue) {
value = anotherValue;
}
@Override
public String toString() {
return value.toString();
}
@Override
public boolean equals(Object obj) {
return value.equals(obj);
}
@Override
public int hashCode() {
return value.hashCode();
}
}
testcase:
public void changeRef(Ref<String> ref) {
ref.set("bbb");
}
// ...
Ref<String> ref = new Ref<String>("aaa");
changeRef(ref);
System.out.println(ref); // prints "bbb"
Can I pass parameters by reference in Java?
No.
Why ? Java has only one mode of passing arguments to methods: by value.
Note:
For primitives this is easy to understand: you get a copy of the value.
For all other you get a copy of the reference and this is called also passing by value.
It is all in this picture:
Source: Stackoverflow.com