Finalize()

[suma krishnan]

Finalize() can be overloaded and can't be overriding.
Correct me if I'm worng

[Ragu Sivaraman]

Originally posted by suma krishnan:
Finalize() can be overloaded and can't be overriding.
Correct me if I'm worng

Finalizer cant be overloaded but can be overridden
ragu

[Anonymous]

Overloading the finalize() method is allowed, but only the method with the original finalize() signature will be called by the garbage collector. Moreover,the new definition cannot contradict the original definition.
As far as overriding is concerned, the finalize() method is declared protected in Object. Therefore, the overriding methods can be only protected or coderanch.
Hope this Helps,
Shyam

[Ragu Sivaraman]

Originally posted by Shyamsundar Gururaj:
Overloading the finalize() method [b]is allowed, but only the method with the original finalize() signature will be called by the garbage collector. Moreover,the new definition cannot contradict the original definition.
As far as overriding is concerned, the finalize() method is declared protected in Object. Therefore, the overriding methods can be only protected or coderanch.
Hope this Helps,
Shyam[/B]

If the new overloaded finalizer() method cannot contradict the original finalizer() how can you overload this method?
protected void finalizer() //
protected int finalizer() // Not a overloading
protected void finalizer(5) //Different Signature so overloading
//But it wont compile!!!
So i dont think you can overload a finalizer...
Please correct me if i am wrong
Thankx Shyam
ragu

[Anonymous]

Ragu,
Actually it can be overloaded. I couldn't believe it at first, but it came from a rather credible source...Mughal..So, I tested out one of the programs in his book...it works!.
Run this code..you'll believe too!

<font size=2>
class SuperBlob {                                     // (1)  
    static int idCounter;
    static int population;
    protected int blobId;
    public SuperBlob() {
        blobId = idCounter++;
        ++population;
    }
    protected void finalize() throws Throwable {      // (2)
        super.finalize();
        --population;
    } 
}
class Blob extends SuperBlob {                        // (3)
    int[] fat;
    public Blob(int bloatedness) {                    // (4)
        fat = new int[bloatedness];
        System.out.println(blobId + ": Hello");
    }
    protected void finalize() throws Throwable {      // (5)
       System.out.println(blobId + ": Bye");
        super.finalize();
    }
    protected int finalize(int value){   // [B]Overloaded Finalizer[/B]
       System.out.println("Overloaded Finalizer");
       return 5;
    }

}
public class Finalizers {
    public static void main(String args[]) {          // (6)
        int blobsRequired, blobSize;
        try {
            blobsRequired = Integer.parseInt(args[0]);
            blobSize      = Integer.parseInt(args[1]);
        } catch(IndexOutOfBoundsException e) {
            System.err.println("Usage: Finalizers <number of blobs> <blob size>");
            return;
        }
        for (int i=0; i<blobsRequired; ++i) {         // (7)>
            new Blob(blobSize);
        }
        System.out.println(SuperBlob.population + " blobs alive"); // (8)
        System.out.println(new Blob(33).finalize(5)); // (8)
    }
}

Waiting for your comments
Shyam

[Ragu Sivaraman]

Thankx shyam that helps
Now my question is Why would somebody overload finalizer in the first place ?
GC only calls the protected void finalizer() signature alone rite?
Ragu

[Anonymous]

You are absolutely right...No one would ever want to overload the finalizer() method. May be Mughal has mentioned this just as a technical point.
Yes! GC calls the original declaration "protected void finalizer() throws throwable".
Hope this helps
Shyam
[This message has been edited by Shyamsundar Gururaj (edited November 01, 2001).]