question with GC

[ved dixit]

I am preparing for SCJP 1.6
I came across this question from the link someone has posted in the forum of java ranch.
The link is here: http://enthuware.com/forum/viewtopic.php?f=16&t=4
I have a doubt for a question related to GC .


How many objects will be eligible for GC just after the method returns?

  


public void compute(Object p)

{

 Object a = new Object();

 int x = 100;

 String str = "abc";

}

 

Ans : 1 .

Please explain the reason.



Thanks

[Neha Daga]

object refered by 'a' is the only object in the method. 'x' is a variable of type int and 'str' refers to 'abc' which is string literal and will be in the string pool.
so there is only 1 object eligible for gc.

[ved dixit]

Hi Neha !!

Thanks for the reply.
So the Object a will be eligible for GC
on line 9 or at the end of the method on line 14 or
when the main method ends (Suppose i have called it from main())?

Thanks.

[Neha Daga]

It will be available for gc at the end of method on line 14.

[ved dixit]

Thankss...i got it !!

[Shanky Sohar]

Hey ,but i studied that String is also an object

[Abimaran Kugathasan]

Shanky Sohar wrote:Hey ,but i studied that String is also an object

Shanky, it's true, that String is also an object. But, here, it's created as String literals on the Pool. So it's not eligible for GC!

[Shanky Sohar]

correct..Abimaran..

[Bert Bates]

on the real exam String objects are never used in GC questions

[Mohit G Gupta]

why object a would be Gc ??
there exists a refernece to object a on line 8.

i think it would be GC when function is no longer needed...

[Wouter Oet]

GC basically happens when an object has no more reference to it.
The object created on line 8 has only 1 reference and that one goes out of scope end the end of line 14.
So after that it's GC'ed.

[Trivikram Kamat]

Bert Bates wrote:
on the real exam String objects are never used in GC questions

But just out of curiosity, if Line 12 is replaced by:

String str = new String("abc");

will there be two objects available for GC after the method returns?

[Neha Daga]

I think yes.

[Abimaran Kugathasan]

Trivikram Kamat wrote:
But just out of curiosity, if Line 12 is replaced by:

String str = new String("abc");

will there be two objects available for GC after the method returns?

I think, the object on the heap will eligible for GC. And the String Literal Pool object won't GCed!

[Trivikram Kamat]

Just read on ScjpFAQ about eligibility of String Literals for GC
It doesn't mention about String objects though.

[Trivikram Kamat]

From Page 434 on K&B

SCJP6 Study Guide wrote:
String s = new String("abc"); // creates two objects,
// and one reference variable

In this case, because we used the new keyword, Java will create a new String object in normal (nonpool) memory, and s will refer to it. In addition, the literal "abc" will be placed in the pool.

This means if we put String str = new String("abc");, there will be two objects available for GC - Object of class Object and that of String.

[Abimaran Kugathasan]

Trivikram Kamat, did you go through that Faq? They have mentioned that,

The objects created in the String pool are not subjected to GC until the class is unloaded by the JVM. They get discarded at that time or just before the virtual machine unloads the class' bytecode.

So, the object, which is created on the literal pool, isn't eligible for normal GC but for VM shut down, etc...

[Trivikram Kamat]

Hi Abimaran Kugathasan,

I agree with you, according to the explanation given in the FAQ on String Literal GC
But it's little confusing, when it says objects created in the String pool are not subjected to GC.

I guess String pool contains literals, and not objects.
Does String str = new String("abc"); creates object in the String pool memory?

According to K&B (as quoted in previous post), Java should create a new String object in normal (nonpool) memory and not in String pool.
However, the literal "abc" is placed in the String pool.

So, the String Object in nonpool memory should be available for GC after the method returns.
However, the literal "abc" will be managed by String pool memory management.

[Trivikram Kamat]

I wrote a small code for testing String objects:

class StringPoolTest {
	public static void main (String[] args) {
		String str1 = "Javaranch";
		String str2 = "Javaranch";
		System.out.println(str1 == str2);
		
		String str3 = new String("Javaranch");
		String str4 = new String("Javaranch");
		System.out.println(str3 == str4);
	}
}

The line 5 prints true.
I guess the reason is both variables are referring to same literal in the String pool.

The line 9 prints false.
I guess the reason is str3 and str4 are two different objects in nonpool memory, both of which contain literal "Javaranch" which is in string pool.

So, objects str3 and str4 should be available for GC at Line 10
Please correct me if I'm wrong.
Thanks.

[Abimaran Kugathasan]

That's correct Trivikram Kamat, but, let's remove the 3,4 and 5 th lines in your above code.

Then, there will be 3 objects totally. One is in the String Constant Pool and other two are in the heap. [String Constant Pool is a special heap, leave it for a while]
Then at the 10th line, the two objects referred by str3 and str4 are only eligible for GC. Not the String Constant Pool objects even though it doesn't have explicit references.

[Trivikram Kamat]

Thanks Abimaran for your quick reply.
So, in the original question if we use String str = new String("abc");, the new String Object created on nonpool memory will be available for GC.

But, I still have one doubt.

Abimaran Kugathasan wrote:
Then, there will be 3 objects totally. One is in the String Constant Pool and other two are in the heap.

The one on the String constant pool ("Javaranch") is a literal, and not an object right?

[Abimaran Kugathasan]

Trivikram Kamat wrote:
The one on the String constant pool ("Javaranch") is a literal, and not an object right?

I think, in Java, other than primitives, all are O(o)bjects. So, that's also an object!

[Trivikram Kamat]

Abimaran Kugathasan wrote:
I think, in Java, other than primitives, all are O(o)bjects. So, that's also an object!

I'm not sure whether a literal can be called an object or not...

The definition of Literal from wikipedia says that:
A literal is a notation for representing a fixed value in source code.

In Oracle Java SE tutorials website, literals is explained under Primitive Data Types.

[Abimaran Kugathasan]

Trivikram Kamat wrote:
In Oracle Java SE tutorials website, literals is explained under Primitive Data Types.

That isn't about the String Constant pool literals. For example true and false are the literal of boolean data type. That's different! Have a look on this JLS

String Constant Pool Literals are String and hence Objects!