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I doubt my garbage collection

 
Gary Marshall
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How many objects are eligible for gc by the time the f() method is invoked?
I say there is only one eligible for gc: "a", since reference variable "a" has been assigned to object b and reference variable "b" has been assigned to object "c", which leaves the object originally referred to by reference variable "a" without a reference. The answer given to me is that the two objects orginally referred to by "a" and "b" are eligible for gc.

Whose correct here? If they're answer is correct could you please help me understand how it is that both objects originally referred to by "a" and "b" are eligible for gc?
Thank you
g
 
dolly shah
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X a = new X();
X b = new X();
X c = new X();
a=b;//line 1
b=c;//line 2
f();

Here at line 1 a is now referring to b. At line 2 b is now referring to c. Both are now referring to c. So both are eligible for GC.
 
ahmed yehia
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At line 2 b is now referring to c. Both are now referring to c

When an Object has more than one reference variable, reassigning one to refer to diff object doesnt reassign the other.
In the example only one object is eligible for GC.
 
Lucky J Verma
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Only 1 should be eligible for GC.
a is pointing to what b pointed
b is pointing to what c is pointing.


what is the correct answer.
 
Gary Marshall
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I agree with Ahmed and Lucky.... reference "a" changed to point to "b", and "b" shifted to point to "c". The result that see is the following:
1. object originally referred to by "a" has no reference, therefore
eligible for gc
2. object originally referred to by "b" is being reffered to by "a", so
there is a reference pointing to this object and therefore this
object is not elgible for gc
3. object originally referred to by "c" is being refrered to by "b"
and "c" so there are two references pointing to this object and
therefore this object is not eligible for gc

As I see it this leaves 1 (one) object eligible for gc.

Can another java guru please help?
thank you
g
 
dolly shah
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Can someone please explain me the difference here in following 2 scenarios?

2 separate situations are here for 3 objects a,b,c. Assume that all three objects are created.

1.

a=b;
b=c;

2.
a=b=c;

So finally in both assignment statements , how many objects will be eligible for GC?
 
Gary Marshall
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Dolly:

Why is this so confusing?

In a=b=c..... a holds the reference for b, which holds the reference for c... So both a and b are pointing to c (if I can use the word "point" as a matter of helping me understand the concept) Correct or no?
thanks
g
 
Gautam Pandey
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The object which a refers to after the lines (meaning now a is pointing to b's object and both b and c are pointing to c's object) is not referenced by any variable or reference. c's not changed.
so i think only ONE object (originally referred to by a)is eligible. I don't see any ambiguity here.
[ September 24, 2007: Message edited by: Gautam Pandey ]
 
Vas Golla
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Iam having a doubt that are a,b and c are references or instances?
 
ahmed yehia
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Here "a" is a reference variable of an instance of this object.
 
Sunil krishnatry
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Only the object pointed by a will be eligible for garbage collection as this is the only object without reference.

Object b and c have are still reachable objects.
 
fred rosenberger
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Can someone please explain me the difference here in following 2 scenarios?

I'll take a stab. assume in each case that originally a refers to 1, b refers to 2, and c refers to 3
1.

a=b;
b=c;

after the first line, a now refers to the same spot as b. so, both a and b refer to 2. c still refers to 3, and nothing refers to 1 anymore.

you run the second line. you change what b refers to, making it refer to the same thing as 3. so, a still refers to 2, b now refers to 3, and c refers to 3, with nothing referring to 1.


2.
a=b=c;

here, because of the associativity of the "=", you can think of this as if it were written

a=(b=c)

so here, first we change b to refer to c. so that would make a refer to 1, b and c both refer to 3, and nothing refers to 2. now, the value returned from the assignment is the thing being assigned, so "(b=c)" becomes, effectively, "c". this then effectivly gives you a=c.

so, now a refers to 3 as well, and nothing refers to 1 or 2.


So finally in both assignment statements , how many objects will be eligible for GC?

in the first example, 1 object. in the second, 2.
 
Gary Marshall
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Well it seems we have difference of opinion amongst us. Dolly maintains that there are 2 objects eligible for gc, while others, including myself, maintain that there is only 1 object eligible for gc.

FYI: the mock exam tesing software says that there are 2 objects eligible for gc, but I continue to doubt their answer.

Anybody else want to doubt with me or try to explain to me why there are 2 objects eligble for gc and not 1?
thanks again
g
 
Paul Clapham
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I think some less confusing explanations could be provided.

First of all, calling the three objects "a", "b", and "c" is confusing because that gets you mixed up with the names of the three variables that initially refer to them. So let's call them X1, X2, and X3.

Initially a -> X1, b -> X2, and c -> X3.

Then after a=b, a -> X2, b -> X2, and c -> X3.

Then after b=c, a -> X2, B -> X3, and c -> X3.

So there are references to X2 and X3, but no references to X1.
 
Gary Marshall
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OK! Well I don't know about the rest of you folks but for me I have to thank our bartender heroes for straigtening this mess out.

So now I see that my answer is correct as the code was structured thusly:

a=b;
b=c;

So this leaves only one object eligible for gc. Had the code stated:

a=b=c;

then I can see that there would be two eligible for gc.

thank you bartenders... I'll have another cup of java..
g
 
dolly shah
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Thank You bar tenders. I got it now.
 
Burkhard Hassel
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Howdy,

you can check for yourself the garbage collector if you override the finalize method an put some output inside.

Example:

Try it.

Yours,
Bu.
 
PK Ghosh
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I tested the code for "a=b=c". Its 2 object garbage collected.

a points to b, hence a is gone
b points to c, hence b is gone and immidiately a is updated its reference to b (because b is alived through c) which is basically points to c.

hope this helps.
 
Gautam Pandey
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Nice Code

Output is

main started
Y-a finalizing
main ready
 
Sean Nguyen
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Hi,
Here is a little code to test both question:
a=b
b=c

and a=b=c

For the first case, 1 object is for GC. For the second case, 2 objects because (b=c is perform first, then a=b is perform later) the right side of the equal is perform first.




import junit.framework.TestCase;



public class Exame extends TestCase {
public void testGC(){
Xname a = new Xname("A");
Xname b = new Xname("B");
Xname c = new Xname("C");
a = b;
b = c;
System.out.println(a.getName() + b.getName() + c.getName());


a = new Xname("A");
b = new Xname("B");
c = new Xname("C");
a=b=c;

System.out.println(a.getName() + b.getName() + c.getName());
}


private class Xname {
private String name;
public Xname(String name){
this.name = name;
}
public String getName() {
return name;
}
}
}
 
Sean Nguyen
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Forgot adding the reulst:

First case;
BCC

Second case:
CCC
 
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