Round Up question re exception& arg type

[Jerry Bustamente]

Good Morning,

I was wondering if anyone can clarify something for me. I do not believe I am violating the copyright rule which I thoroughly support. Please correct me if I am wrong.

There is a Round Up question that reads, "An exception can be caught by a catch block with an argument type that is a superclass of the thrown exception."

The answer is: "If there is no catch statement that matches the thrown exception , then an argument which is a superclass of the exception will catch it."

Frankly, I just don't understand the answer. The word "argument" in particulary is throwing me.

Can someone explain this to me?

Thanks alot!

:-)

Jerry

[Corey McGlone]

When you have a catch block, you provide an "argument" like this:

try
{
    ...
}
catch (IOException ioe)
{
    ...
}
catch (Exception e)
{
    ...
}

In this example, "IOException ioe" and "Exception e" are the "arguments" that are being passed to the exception block.

Say, for example, that our try block throws a RuntimeException, which is a subclass of Exception. There is no catch block for a RuntimeException. However, the way exception handling works is that, if no catch block is found for the given type of exception, the JVM will look for a catch block of any superclasses of the exception thrown. In this case, we have one - Exception. Because a RuntimeException "is a" type of Exception (it extends Excepetion), this makes sense. Therefore, since there is no specific way to handle a RuntimeException, we'll handle it just like we would an Exception.

I hope that helps,
Corey

[Gian Franco]

Hi,

The argument type is a synonym for parameter
type in the case of try/catch blocks just
like in a method's signature.

Suppose there is only a catch(Exception e)
block and in the try block an IOException is thrown,
then then that catch block will be entered.

Cheers,

Gian Franco

[Jerry Bustamente]

Thank you both for your prompt replies. I understand now.

:-)

Jerry Bustamente