References, objects, passing by value, etc.

Question

This question is more for peace of mind than trying to fix a program. I am trying to understand Java more.
Some simple code:
LinkedList<String> L1;
void myMethod()
   String S1 = "";
   L1.add(S1);
}So I have a LinkedList<String> called L1. I also have a method myMethod(), in which I initialize a String object, S1. I want to call the add() function to add S1 to L1. What happens to the String?
When myMethod exits, the reference S1 is destroyed. Since S1 was copied as a parameter to add(), L1 now has its own reference to the String that was pointed to by S1. So is the String itself destroyed when myMethod exits, thus making LinkedList<String> a totally useless class? Apparently not. That means an object survives as long as there is a reference to it, even if the method in which it was created has exited. Correct?
I have another question, but String is an immutable object so I am thinking what I can use instead... let's use Point.
LinkedList<Point> L1;
Point myMethod()
   Point P1 = new Point(2, 3);
   L1.add(P1);
   return P1;
P1.set(3,4);
System.out.println(L1.getLast());
What does it print?
Many thanks,
daarklord

Answer 1

daarklord wrote:
That means an object survives as long as there is a reference to it, even if the method in which it was created has exited. Correct?Yes. As long as an object has at least one reference to it, it cannot be GC.
LinkedList<Point> L1;
Point myMethod() {
    Point P1 = new Point(2, 3);
    L1.add(P1);
    return P1;
P1.set(3,4);
System.out.println(L1.getLast());For starters this will generate an error since the P1 variable is declared inside the method and is not available outside the method. However if we ignore this fact then since the P1 variable and the List both hold a reference to the same object, if you make a change then both references will reflect that change.