How does Java keep the value from temporary variable?

Question

Hello all:
I have a Question for Java Assignment operator.
Object v1;
Object v2;
v1 = v2; // Java assignment by reference if I understand correctly
///// for example:
import java.io.*;
import java.util.*;
public class TestOne {
Vector v1;
public void changeV() {
Vector v2 = new Vector();
v2.addElement(new Integer(10));
v2.addElement(new Integer(11));
v2.addElement(new Integer(12));
v1 = v2;
public TestOne() {
changeV();
System.out.println(v1.size());
System.out.println("0: " + v1.elementAt(0));
System.out.println("1: " + v1.elementAt(1));
System.out.println("2: " + v1.elementAt(2));
public static void main(String[] args) {
new TestOne();
// Output from Screen
bash-3.00$ java TestOne
3
0: 10
1: 11
2: 12
If I understand correctly, " v1 = v2; " is assigned by reference.
So the question is when function 'changeV' return, why v1 still holds the
values from v2 which has been destroyed?
Thank you
-daniel

Answer 1

firstly, the assignment is by-value, not by-reference.
the value the variables hold is a thingy called "reference." (In other languages such as C or C++ it might be called "pointer.") So "v1 = v2" always copies the value of v2 to v1.
secondly, the vector object that is created in the changeV method, is created on the heap, that is, in dynamic memory. The variable v2 holds only a reference to it. So when the method returns, only the reference is destroyed, not the actual object.
The object is automatically destroyed when there are no more reachable references to it (it is "garbage collected").