Throw difference exception in constructor

Question

Have a base class constructor that throws an exception of class XException
Have an exception class YException that inherit exception XException
If I inherit the base class, can I make it throw exception Y instead of X, since Y is a subclass of exception X?
Example:
public class Base {
public Base() throws XException {
public class XException extends Exception {
public XException() {
// etc...
public YException extends XException {
YException() {
super();
public MyClass extends Base {
// Can this work in a constructor since YException is a subclass of XException?
// Anyway to make this possible?
public MyClass() throws YException {
super();

Answer 1

Have a base class constructor that throws an exception
of class XException
Have an exception class YException that inherit
exception XException
If I inherit the base class, can I make it throw
exception Y instead of X, since Y is a subclass of
exception X?
Example:
public class Base {
public Base() throws XException {
public class XException extends Exception {
public XException() {
// etc...
public YException extends XException {
YException() {
super();
public MyClass extends Base {
// Can this work in a constructor since YException is
a subclass of XException?
// Anyway to make this possible?
public MyClass() throws YException {
super();
Apart from what you are asking in normal circumstances a no args constructor throwing an exception is never a good idea. The reason being that from then on anybody who extends this class would have to face problems. Consider the following
public class MyTest
      public MyTest() throws Exception
class Test extends MyTest
{}this would not compile and throw the following error
Default constructor cannot handle exception type Exception thrown by implicit super constructor. Must define an explicit constructor     
So thats not a good practice
cheers