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Islands of object..

 
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Hi, I got it from K&B 5 study guide and i modified a bit.
class Island
{
Island i;
public static void main(String args[])
{
Island i1=new Island();
Island i2=new Island();
Island i3=new Island();
Island i4=new Island();
i1.i=i2;
i2.i=i3;
i3.i=i1;
i4.i=i3;
i3=null;
i2=null;
i1=null;
}
}

How many objects are eligible for GC?
my guess is 3. since i4 is not null islands of objects are still referring each other and they are not eligible for Gc.so only i1,i2,i3 are eligible for Gc.
Am i got the point?

Please do correct me if i am wrong.

Preparing Scjp 1.5
 
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only i1 and i2 are eligible for GC.
i3 is being referenced by i4.i !
 
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No objects are eligible for GC. From the i4 reference you can get to the i4 object. From the reference in that object, you can get to the object originally referred to by i3. From the reference in that object, you can get to the object originally referred to by i1. And from the reference in that object, you can get to the object originally referred to by i2.

Henry
 
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I agree with Henry.

To make a cheap diagram you can still access all objects in the following way:

i4->i4.i->i3->i3.i->i1->i1.i->i2

i2 then references i3 but since we already accessed that from i4 it doesn't matter. All objects can be referenced through i4 so none qualify for GC.
 
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yes no objects are eligible for GC.

i1----------object1-----------i3.i

i2---------object2-------------i1.i

i3---------object3-------------i4.i
|
|
i2.i


i4---------object4

when i3=null


i1----------object1-----------i3.i

i2---------object2-------------i1.i

object3-------------i4.i
|
|
i2.i


i4---------object4

when i2=null


i1----------object1-----------i3.i

object2-------------i1.i

object3-------------i4.i
|
|
i2.i


i4---------object4

when i1=null

object1-----------i3.i

object2-------------i1.i

object3-------------i4.i
|
|
i2.i


i4---------object4


so no object is eligible for GC.....
[ November 11, 2008: Message edited by: Ganeshkumar cheekati ]
 
Preethi Dev
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thanks to all i am clear now.
 
M Srilatha
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I made a mistake

Thanks to all!
 
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