Garbage Collection Eligibility

[Suhrid Karthik]

Here's a question from Khalid Mughal's SCJP book:

How many objects are eligible for garbage collection when control reaches (1) ?

public class Eligible {
   public static void main(String[] args) {
            for(int i=0; i < 5; i++) {
                Eligible obj = new Eligible();
                 new Eligible();
           }
           System.gc();      // (1);
}

Select the one correct answer:

(a) 0
(b) 5
(c) 10
(d) Hard to say.

My answer was (c) 10.
But the book says the answer is (d) Hard to say. The reasoning is that some of the object's could have been GC'ed when control reaches (1).
I see the point - some objects could have been automatically GC'ed before we explicitly call GC.

My question is that is this a valid way to think about GC questions on the real exam ? I would assume that ten objects will be eligible for GC as soon as the loop ends.

[John Stark]

new Eligible();

generates an anonymous Eligible object which is immediately eligible for garbage collection. After the first time the loop body has been executed

Eligible obj = new Eligible();

removes the only reference obj was pointing to before, making the old Eligible object immediately eligible for garbage collection.
So 9 objects are eligible for gc even before the for loop finishes. How do you know garbage collection at least of some objects has not already taken place when

System.gc();      // (1);

is reached? But, yes I was also puzzled whether they want to hear '10' or 'hard to say'.

John

[Stephan van Hulst]

I'm afraid you're mistaken John. At the end of every iteration of the loop, 2 more objects will be eligible (assuming no GC takes place in the mean time).

The variable obj goes out of scope after the iteration, so the reference is lost even before it gets overwritten with a new object.

[Scotty Mitchell]

I agree with stephen in this situation. The obj reference variable is lost after each iteration, hence you arnt sure what has been already GC or what needs to be at the end of the for loop.

[John Stark]

That's right, sorry.

John

[Suhrid Karthik]

Thanks everyone. The book's answer makes sense now.