Re: Exception Handling
Jeff KE7NVY wrote:
On Mar 9, 9:53 pm, Javas <deepan...@gmail.com> wrote:
public class Qn24 {
static class A
{
void process() throws Exception { throw new
Exception(); }
}
static class B extends A
{
void process() { System.out.println("B "); }
}
public static void main(String[] args)
{
A a = new B();
a.process();
}
}
Why does the above code results in a compilation error? since it is
always invoking class B's process() method which doesnt throw any
exception.
Exception is a checked exception. That means that whenever you
declare
a method that (potentially) throws it, you must do one of two things
in
the code that calls that method. Either handle the exception by
putting
the call inside a try/catch block:
public static void main(String[] args) {
A a = new B();
try {
a.process();
}
catch (Exception e) {
System.out.println("main: Exception caught");
}
}
I don't like code that catches Exception and neither re-throws it,
possibly wrapped with additional data, nor terminates the program. If
this main method were called from another class, rather than being used
as the top method of the whole application, the caller could be left
continuing to execute with the processing not done and no indication
that something went wrong other than a line in standard out.
Or, declare the calling method as also throwing Exception:
public static void main(String[] args) throws Exception {
A a = new B();
a.process();
}
A third option is to change the declaration to "B a = new B();". The
compiler could then use the knowledge that variable a refers to a B
instance.
Patricia