Thread: std::bad_exception don´t throw exception

Hmmm, can´t really grasp the logic of std::bad_exception or I am simple using it wrong. I´ve tried to make a board seach on the topic but I wasn´t succesful. If I understand it correctly std::bad_exception is used when an exception is thrown that isn´t declared in the exception specification, e.i.

Code:

#include <iostream>
#include <conio.h>
#include <exception>

using namespace std;

class A
{};
class B
{};

void test() throw(A, std::bad_exception)
{
    throw B();//Here should it throw a std::bad_exception type 
//throw A(); //This works okey 
}

int main(void)
{

    try
    {
        test();    
    }
    catch(A)
    {
        cout << "Error in CLASS A" << endl;
    }
    //Can´t catch it here
    catch(std::bad_exception)
    {
        cout << "Error in std::bad_exception" << endl;
    }

    getch();

return 0;
}

When I run this code I get a "abnormal program termination". When I delved deeper in my book I found that that the line

Code:

throw B();//Here should it throw a std::bad_exception type

actually doesn´t throw it directly but it goes through a function called unexpected() which calls another function, terminate(), that terminates the program (like in my case). Well if it terminates the program and you have no acces to the "supposed" thrown object, then whats the purpose of std::bad_exception??? I´m totaly lost.