Need help on this piece of code about GC and finalize()

[Jayes Herryl]

Hi, I tried to run this piece of code:
public class Test{

public static void main(string arg[]){
Object obj = buildTest(3);
System.gc();
System.out.println("Existing");
}

public static test buildTest(int n){
Test t = null;
for (int i = 0; i < n; i++){
t = new Test(i);
}
return t;
}
string name;
public Test(int n){
name = "Number " + n;
}

public void finalize(){
System.out.print("finalize " + name);
}
}

The output I got was:
finalize Number 1
finalize Number 0
Existing

My questions is, why it does not print out "finalize Number 2"? If I pass 4 to buildTest in main(), it will not print "finalize Number 3" as well... Can anyone explain this? Is it because of the first line "Test t = null;" in buildTest()?
Any help is appreciated!!

[ June 09, 2006: Message edited by: Jay Cat ]
[ June 09, 2006: Message edited by: Jay Cat ]

[wise owen]

Object obj = buildTest(3);
The buildTest(3) returns the object ("Number 2") to caller and assign to variable obj.

[Firas Zuriekat]

It's because obj is referring to an Test object. It keeps referring to it all the way until the left brace "}" in main () is reached.

But setting obj to null as follows:

public class Test{
public static void main(String arg[]){
Object obj = buildTest(3);
obj= null;
System.gc();
System.out.println("Existing");
}
static public Test buildTest(int n){
Test t = null;
for (int i = 0; i < n; i++){
t = new Test(i);
}
return t;
}
String name;
public Test(int n){
name = "Number " + n;
}
private Test(){}

public void finalize(){
System.out.println("finalize " + name);
}
}

gives OUTPUT:

finalize Number 2
finalize Number 1
finalize Number 0
Existing

[Jayes Herryl]

Now I see why...thanks a lot for the explaination!