3

For the record, I'm NOT a Java Beginner, but -- rather - an intermediate-level guy who kinda forgot a bit about fundamentals about Java.

class C{
    public static void main(String a[]){
        C c1=new C();
        C c2=m1(c1);        //line 4
        C c3=new C();
        c2=c3;              // line 6
        anothermethod();
    }
    static C m1(C ob1){
        ob1 =new C();      // line 10
        return ob1;
    }
    void anothermethod(){}
}

From the above code:

  • Why is it that after line 6, 2 objects of type C are eligible for Garbage Collection(GC)?

  • Why isn't it that in line 4 and line 10, a copy of c1 is passed into the m1() method. So, eventually in line 6, there is going to be 1 Object (not 2) that are going to be eligible for GC. After all, isn't java pass -by-value rather than pass-by-reference?

Catfish
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2 Answers2

4

There's a difference between pass-references-by-value and pass-values-by-reference :)

Is Java Pass By Reference
Java is never pass by reference right right
Pass By Reference Or Pass By Value

You might want to check out Jon Skeet's article on C# parameter-passing semantics as well, seeing as it's his favorite 'programmer ignorance' pet peeve:
What's your favorite 'programmer ignorance' pet peeve.

So basically, I see your code do the following:

c1 = new C("Alice");
    // m1(C obj1) {     -- c1 gets passed to m1, a copy of the reference is made.
    //                  -- there are now two references to Alice (c1, obj1)
    //    obj1 = new C("Bob"); -- there is now one reference to Alice
                                // and one reference to Bob
    //    return obj1;  -- returns a reference to Bob(c1 still reference Alice)
    // }                -- when m1 returns, one of the references to Alice disappears.
c2 = m1(c1); // c2 points to Bob 
c3 = new C("Charlie");
c2 = c3;      // <-- Bob is eligible for collection. 
              // There are now two references to Charlie
Community
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Jimmy
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1

What makes you think two objects of type C are available for GC after line 6? I only see one (c2). What tool are you using that tells you otherwise?

Regarding your question about passing c1 into your m1 method: What you pass (by value) is a reference to the object -- a handle by which you can grab the object, if you will -- not a copy of it. The fact you pass a reference into m1 is completely irrelevant, in fact -- you never use that reference, you immediately overwrite it with a reference to a new object, which you then return (this does not affect the c1 that is still referenced in main).

T.J. Crowder
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  • Refer to question 13 in the following link: http://www.examsguide.com/scjp/freequestions2.html – Catfish Feb 19 '10 at 16:53
  • @Catfish: We're not the only ones who think they're wrong: http://www.coderanch.com/t/452512/Programmer-Certification-SCJP/certification/help-garbage-collection – T.J. Crowder Feb 19 '10 at 17:18
  • @Catfish: With respect to them, they're wrong. As of Line 6 in your code (and it looks about the same as theirs), the first C created (by main) is still referenced as c1 in main; the second C created (by m1) is no longer referenced (because the variable that was referencing it is no referencing the third one) and may be GC'd; the third instance created (by main) is still referenced by both `c2` and `c3` in `main`. Thus, the second object is the only one ready for GC. At least (the only one mentioned in that code). Apologies if you read my earlier breakdown, I got my `c2` and `c3` mixed up. :-) – T.J. Crowder Feb 19 '10 at 17:22