C++ Class Question

[sake]

Hi,

I recently came about an article on cplusplus.com regarding exception handling and, specifically, a code excerpt:

// standard exceptions
#include <iostream>
#include <exception>
using namespace std;

class myexception: public exception
{
  virtual const char* what() const throw()
  {
    return "My exception happened";
  }
} myex;

int main () {
  try
  {
    throw myex;
  }
  catch (exception& e)
  {
    cout << e.what() << endl;
  }
  return 0;
}

The part that confuses me is after the class myexception is declared.

class myexception: public exception
{
  virtual const char* what() const throw()
  {
    return "My exception happened";
  }
} myex;

The last word of this snippet outside the braces: myex;. What does this mystery word mean and do? After looking at the code, I figured it may have been a shortcut to referring to the class, but it seems silly to have do that when you could just make the class's name "myex". Anyone care to shed some light on this for me?

Thanks,
Austen

[Banfa]

It is declaring a variable named myex of type myexception. The code is equivalent to

class myexception: public exception
{
  virtual const char* what() const throw()
  {
    return "My exception happened";
  }
};

myexception myex;

[sake]

Ah, I see.
Thank you very much.
-Austen