Is there a way to throw an exception without adding the throws declaration?

I have the following situation.

I have a Java Class that inherits from another base class and overrides a method. The base method does not throw exceptions and thus has no throws ... declaration.

Now my own method should be able to throw exception but I have the choices to either

• Swallow the exception or
• Add a throws declaration

Both a not satisfying because the first one would silently ignore the exception (ok I could perform some logging) and the second would generate compiler errors because of the different method headers.

public class ChildClass extends BaseClass {

        @Override 
        public void SomeMethod() {
            throw new Exception("Something went wrong");
        }
}

Solution 1:

You can throw unchecked exceptions without having to declare them if you really want to. Unchecked exceptions extend RuntimeException. Throwables that extend Error are also unchecked, but should only be used for completely un-handleable issues (such as invalid bytecode or out of memory).

As a specific case, Java 8 added UncheckedIOException for wrapping and rethrowing IOException.

Solution 2:

Here is a trick:

class Utils
{
    @SuppressWarnings("unchecked")
    private static <T extends Throwable> void throwException(Throwable exception, Object dummy) throws T
    {
        throw (T) exception;
    }

    public static void throwException(Throwable exception)
    {
        Utils.<RuntimeException>throwException(exception, null);
    }
}

public class Test
{
    public static void main(String[] args)
    {
        Utils.throwException(new Exception("This is an exception!"));
    }
}

Solution 3:

A third option is to opt out of exception checking (just like the Standard API itself has to do sometimes) and wrap the checked exception in a RuntimeException:

throw new RuntimeException(originalException);

You may want to use a more specific subclass of RuntimeException.