Why does std::exception catch my exception before std::bad_alloc?

Question

Problem : I am using both std::exception and std::bad_alloc to catch exception. Something is wrong with the order of the try catch that I am using. I attached sample code for reference.

Expected : If my error is bad_alloc then the bad_alloc exception is thrown.

Observed : My error is bad_alloc, but exception is thrown.

Sample Code :

#include "stdafx.h"
#include <iostream>
#include <exception>

using namespace std;

void goesWrong()
{
    bool error1Detected = true;
    bool error2Detected = false;

    if (error1Detected)
    {
        throw bad_alloc();
    }

    if (error2Detected)
    {
         throw exception();
    }
}

int main()
{
    try
    {
        goesWrong();
    }
    catch (exception &e)
    {
        cout << "Catching exception: " << e.what() << endl;
    } 
    catch (bad_alloc &e)
    {
        cout << "Catching bad_alloc: " << e.what() << endl;
    }

    return 0;
}

Answer 1

You have to put your exceptions in reverse order, regarding their inheritance relationship. std::exception is the parent class of std::bad_alloc, that is why it is found before in the catch list. So you have to transform your code to be:

   try {
      goesWrong();
   }
   catch (bad_alloc &e)
   {
      cout << "Catching bad_alloc: " << e.what() << endl;
   }
   catch (exception &e)
   {
      cout << "Catching exception: " << e.what() << endl;
   }

You're not limited to catch objects: you can throw integers, chars... whatever. In that case, catch(...) is the only secure way to catch them all.

That said, using objects from the standard class library is the advised way to do it. And in this case, since std::exception is the base class for all (standard) exceptions, it will catch all possible exceptions thrown.

You can create your own exception classes deriving them from std::exception, or from std::runtime_error, for example, my personal choice.

Hope this helps.