GC Question from Marcus Green

[Paulo Aquino]

public class BettyAck {
    public static void main(String argv[]){
BettyAck b =new BettyAck();
    }
    public BettyAck() {
Integer x = new Integer(10);
        findOut(x);
Integer y = new Integer(99);
        Integer z = y;
        z = null;
        findOut(y); 
        //here
    }
    public void  findOut(Integer y){
y = null;
        
    }
}
 
  

Given the following code, how many objects will be eligible for garbage collection on the line with the comment //here

0
1
2
3
4

I answered 2, but the answer given was 0. Why is this so?

[Pradeep bhatt]

 
Integer x = new Integer(10);//1
        findOut(x);//2
Integer y = new Integer(99);//3
        Integer z = y;//4
        z = null;//5
        findOut(y); //6 

Zero is correct. Line 1 creates a new integer object. In line a copy of refernce x is passed to method. In the method findOut, the copy is reassigned , not the original object. The original object pointed by x stil exists. Line 4 causes one more reference to point to object created at line 3. z =null makes only y point to the object. Apply similar logic as applied previously for findout(y). When the commented line is reached all objects are refernced and are not eligible for GC. Hope this helps

[Samir Arora]

Hi,
According to me also it should be 2 only. Lets see if any one can explain !!

[Manish Nijhawan]

Hi Pardeep,
As you have explained that at the commented line since all the objects have the references so they are not eligible for the GC. But since that line is the last line of the method , so after that line all the references to all the objects will vanish , so all the Live objects should be eligible for GC.

Please explain me this as i am very much confused about GC.

[Srinivasa Raghavan]

I agree none are eligible for GC.

But if the findOut() is modified like the one below then 2 objects are eligible for garbage collection , am i right ?

 
public void  findOut(Integer y)
{
y = new Integer(111);
        y = null;            
}

[ April 21, 2005: Message edited by: Srinivasa Raghavan ]

[Pradeep bhatt]

Originally posted by Srinivasa Raghavan:
I agree none are eligible for GC.

But if the findOut() is modified like the one below then 2 objects are eligible for garbage collection , am i right ?

 
   public void  findOut(Integer y)
{y = new Integer(111);y = null;            }

yes

[Pradeep bhatt]

Originally posted by Manish Nijhawan:
Hi Pardeep,
As you have explained that at the commented line since all the objects have the references so they are not eligible for the GC. But since that line is the last line of the method , so after that line all the references to all the objects will vanish , so all the Live objects should be eligible for GC.

Please explain me this as i am very much confused about GC.

The objects will be eligible for GC after the method completes.

[soumya ravindranath]

Hi,

I understand that the object holding 111 in findOut() becomes eligible for GC, but which is the other object that's eligible after the modified findOut() ?

Thanks,
Soumya
[ April 21, 2005: Message edited by: soumya ravindranath ]

[Srinivasa Raghavan]

findOut() is called twice check the code

public BettyAck() 
{
Integer x = new Integer(10);        
findOut(x);
Integer y = new Integer(99);        
Integer z = y;        
z = null;        
findOut(y);         //here    
}

[soumya ravindranath]

Oooops! thanks

[nishant vats]

According to me only one object will be elgible for garbage collection..i.e
new Integer(10);
this object has refernce count =0;as u can check .Intially this object was refrenced by x but when x=y is executed .This object is no longer refrenced.
new Integer(99) is being still refernced by y.so only one object is elgible for garbage collection

[Pradeep bhatt]

Originally posted by nish vats:
According to me only one object will be elgible for garbage collection..i.e
new Integer(10);
this object has refernce count =0;as u can check .Intially this object was refrenced by x but when x=y is executed .This object is no longer refrenced.
new Integer(99) is being still refernced by y.so only one object is elgible for garbage collection

Where is x=y I dont see it. What about others?

[soumya ravindranath]

As I see it, there is no x =y in the code given.

When z = y and then z = null, as Pradeep said, it means nothing to z or y. x continues to point to 10.

[Paulo Aquino]

Now I understand why "0" was the answer. The copy of the reference object was reassigned in findOut method and not the original reference.

[Sunyaluk Boonmas]

The answer is so clear.

This is the sample of pass by reference problem.

This problem look like this

http://www.coderanch.com/t/248832/java-programmer-SCJP/certification/Dan-exam

This explanation very clear, I think.

http://www.javaranch.com/campfire/StoryPassBy.jsp

And this flash playing is interesting.I think it is very good to show how it work.


http://www.geocities.com/mcglonec1978/javacert/paramPassing_highres.html

This it main url.
http://www.geocities.com/mcglonec1978/javacert/javacert.html#param_passing

Hope it useful for you.