Please help with try/catch

Question

Hello, I'm writing a program and I got surprised with the following. I've got the following try/catch block:
try {
String param = config.getConfigValue("db2.driver");
if (param == null)
// parameter not defined.
throw new Exception("Parameter 'db2.driver' missing!");
else
Class.forName(param);
catch (SQLException sqlEx) {
// do something
catch (Exception e) {
System.out.println("*** ERROR loading driver..."); (1)
finally {
releaseDB2Resources();
I know the program flow goes through the finally block, but it doesn't print the message (1). Why is that? I don't understand it. What am I doing wrong?
Thanks in advance.

Answer 1

A finally block always gets executed, so that's why the println always works there. The only reason why the println in the catch wouldn't work is if that exception wasn't thrown. If a config value is not there, it might be the null string "" instead of null. Try this:
if (param.equals(""))
throw new Exception("blah");
You could also try:
if (true)
throw new Exception("blah");
to make sure that the exception is caught where you think it should be.