Why does this program fail to catch an exception?

Question

I'm trying to print the type name using exceptions, but my program doesn't even seem to catch the exception and instead seems to call the default termination function. What have I missed?

#include <cstdio>
#include <exception>
#include <typeinfo>

namespace Error
{
    template<typename T>
    class Blah : std::exception
    {
        virtual const char* what() const throw()
        {
            return typeid(T).name();
        }
    };
}

void blah() {
    throw Error::Blah<int*********>();
}

int main()
{
    try
    {
        blah();
    }
    catch (std::exception& e)
    {
        std::puts(e.what());
    }
}

score 10 · Accepted Answer · answered Apr 19 '15 at 19:35

The problem is here:

template<typename T>
class Blah : std::exception
//          ^^^^^^^^^^^^^^^

You're inheriting privately (since class inheritance is private by default and you don't add a specifier), so std::exception isn't an accessible base. You have to inherit publicly.

Why does this program fail to catch an exception?

1 Answers1