Question about declaring finalize()

[Leonard Fischer]

Hello everyone. I can declare a finalize() method as follows:

protected void finalize() throws Throwable {
out.println("in finalize()");
super.finalize();
}

No problem. But I don't understand why the following does not compile:

protected void finalize() throws Exception{
out.println("in finalize()");
super.finalize();
}

Of course, Exception is "narrower" than Throwable, and I realize that super.finalize() throws a Throwable.

HOWEVER, it seems to me (although I'm not 100% sure) that everything that is a Throwable, and is NOT an Exception, is either an Error, or an Unchecked Exception. And my understanding is that Errors and unchecked exceptions (RuntimeException's) do NOT have to be "handled and declared".

(Even if I'm correct, I wouldn't think of this as a compiler bug... but I'm curious... I expected "throws Exception" to compile OK.)

Thanks!

[Ruben Soto]

At some point I was surprised by this too. It turns out that Throwable is a wider checked "exception" than Exception. I think the idea is that you could subclass Throwable directly, and have a checked "exception" hierarchy which wouldn't be a descendant of Exception, but as long as it is not a descendant of RuntimeException, it would be checked. This is just my interpretation of why this works this way though, and it might be incorrect.

The bottom line is that if you call a method which throws Throwable, you either have to catch Throwable, or declare that you throw a Throwable. It's not enough with catching or declaring Exception.

[Vinod Kumar Kommineni]

Hi Leonard

Throwable can be either checked or uncheked exception

and also same is the case with Exception it will be a checked exception unless it belongs to RuntimeException branch ,

so declaring it here like this

protected void finalize() throws Exception{
out.println("in finalize()");
super.finalize();
}    

is basically violating the rule that : new or wider checked Exception that are not thrown by super class version of a method should not be thrown by its overriding versions

because when you are calling the super class version it might throw a Throwable which is super class to the class Exception (Since it's a super class version , it can throw a wider Exception)

so modifying the code to handle it will solve the issue:

 protected void finalize() throws Exception{
		System.out.println("in finalize()");
		try {
			super.finalize();
		} catch (Throwable e) {
			
			e.printStackTrace();
		}
		}    

or else if you modify the super class method implementation and make sure it throws Excepetion

it should work i guess.

for Throwable there is no problem because when you are saying

    protected void finalize() throws Throwable {
out.println("in finalize()");
super.finalize();
}

it should handle it or the method calling will handle it but its a Throwable in both cases which is the super class of all errors and exceptions in the Java language.
so its not showing any compilation error i guess.

I hope this explanation is correct.

Some one correct me if i am wrong

[Leonard Fischer]

Thanks to Ruben and Vinodkumar. What you say makes sense; someone COULD subclass Throwable to make their own checked exception.