How to calculate eligibility of objects to garbage collector in Java?

Question

Please clarify my doubt.

class A{
Long weight = 1000L;
}
public class B extends A{
 public static void main (String[] args){
    B b = new B();
    B c = new B();
    b = null; 
    c = null;//going to gc.
 }
}

Here in the above code, while reaching "going to gc" how many objects are eligible for garbage collector? As far as i know args[], b, c and two Long objects totally 5 are eligible. But some says totally 4. They are saying that no two Long object will be created in heap but only one.

Please clarify my doubt.

The question is impossible to answer as the JVM is allowed to cache Long objects on it's own accord, not just based on the internal cache in the Long class that caches values from -128 to 127. Of course if you ignore this fact, then Jigar Joshi's answer is correct. — MeBigFatGuy, Apr 26 '11 at 05:53

score 3 · Accepted Answer · answered Apr 26 '11 at 05:43

3

4 Object[2 of B, 2 of Long] will be created and all 4 would be ready for GC.

At the line

c = null;//going to gc.

there would be live reference of args so args won't be ready

answered Apr 26 '11 at 05:43

jmj

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Thanks for your response. So. After main() executes there will be 5. Am I right? – Manoj Apr 26 '11 at 06:37

Edwin Dalorzo · Answer 2 · 2011-04-26T06:03:54.440

According to the Java Language Specification section 5.1.7 Boxing Conversions:

If the value p being boxed is true, false, a byte, a char in the range \u0000 to \u007f, or an int or short number between -128 and 127, then let r1 and r2 be the results of any two boxing conversions of p. It is always the case that r1 is the
same as r2.

Therefore it may be true only if this principle applies to long autoboxing of 1000L as well, false otherwise.

This could be the premise of the question, but JLS does not seem to include long autoboxing, neither numbers over 127,

Chances are that you are right an your questioner is wrong as far as I can see.

At any rate, I just wanted to point out that the scenario in question is possible with autoboxing under the conditions stated in the JLS. The conditions on your question do not satisfy the requirements and therefore I'd agree with your elligible object count.

Longs are also autoboxed in some implementations... Have a look at the implementation of Long.valueOf(long) — Dilum Ranatunga, Apr 26 '11 at 05:50
No There would be two different objects of Long .check [this](http://stackoverflow.com/questions/5581913/wrapper-class-and-operator) — jmj, Apr 26 '11 at 05:51

score 0 · Answer 3 · answered Apr 26 '11 at 05:42

0

Well, if b.weight == c.weight then the answer is 4. Otherwise, the answer is 5. Try it and see.

answered Apr 26 '11 at 05:42

Lyn Headley

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score 0 · Answer 4 · answered Apr 26 '11 at 05:49

This is slightly tricky.

'b', 'c' get garbage collected. +2

Since weight's value is outside of the range [-128, 127], the autoboxing will allocate new Long instances for each 'A' +2 (See http://www.jdocs.com/harmony/5.M5/java/lang/Long.html#823)

If you were invoking B.main using some reflection mechanism, and did not hold references to class B or its Method instances etc, then class A, class B along with all the Method instances etc. +a_bunch

If the invoker does not use the args array passed in, then args; +1

score 0 · Answer 5 · answered Apr 26 '11 at 05:50

There are two Long objects. You can verify by adding these lines before you set b and c to null:

System.out.println(System.identityHashCode(b.weight));
System.out.println(System.identityHashCode(c.weight));

You'll get two different identity hash codes meaning they're two different objects.

score 0 · Answer 6 · answered Apr 26 '11 at 06:03

0

only 4 objects are eligible for gc . at line c = null;//going to gc. set the break point and debug your code in eclipse it will show you how many object will created.

answered Apr 26 '11 at 06:03

p27

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How to calculate eligibility of objects to garbage collector in Java?

6 Answers6