Implement serializable if I already extend it?

[timasmith]

If I extend an exception (which can be serialized) do I need to
explicitly implement serializable?

public class MyException extends Exception {

static final long serialVersionUID = 1L;

public MyException(Exception exception){
super(exception);
}

}

thank

Tim

 

[zero]

(e-mail address removed) wrote in @f14g2000cwb.googlegroups.com:

If I extend an exception (which can be serialized) do I need to
explicitly implement serializable?

public class MyException extends Exception {

static final long serialVersionUID = 1L;

public MyException(Exception exception){
super(exception);
}

}

thank

Tim

No.

The main reason for mentioning it, even though it's not really necessary,
would be to make it more obvious that you're implementing it. So it's
mostly for readability.

 

[Roedy Green]

If I extend an exception (which can be serialized) do I need to
explicitly implement serializable?

public class MyException extends Exception {

static final long serialVersionUID = 1L;

public MyException(Exception exception){
super(exception);
}

Any subclass automatically implements all the interfaces of the base
class. You can optionally provide another serialVersionUID which can
be considered the version number of the extension fields
..