Why does MinGW 4.8 not treat ex.what() as virtual?

Question

I'm using Qt Creator 3.0.1 with MinGW 4.8 32-bit as the compiler.

When I put the following code into the very top of my main() function (before Qt does all its stuff), the output I get on the console is "std::exception", not "Whoops" as I would expect:

try {
    throw std::logic_error{"Whoops"};
}
catch (std::exception ex) {
    std::cout << ex.what() << std::endl;
}

I've also tried accessing what() through a pointer:

try {
    throw std::logic_error{"Whoops"};
}
catch (std::exception ex) {
    std::exception* ex2 = &ex;
    std::cout << ex2->what() << std::endl;
}

The exact same code compiled in VS2013 outputs "Whoops" as I would expect.

std::exception::what() is virtual, so why is this happening?

See http://stackoverflow.com/questions/274626/what-is-the-slicing-problem-in-c. `ex` is a `std::exception`, both statically and dynamically. — chris, Apr 02 '14 at 19:40

score 1 · Accepted Answer · edited May 23 '17 at 12:11

1

As chris said, you're slicing the information. You can prevent this by catching a const reference instead (demo):

try {
    throw std::logic_error{"Whoops"};
} catch (const std::exception& ex) {
    std::cout << ex.what() << std::endl; // Whoops
}

edited May 23 '17 at 12:11

Community

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answered Apr 02 '14 at 19:44

Zeta

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I just found that out after looking into the link @chris posted above. I'll accept in 7 minutes when SO lets me :) – Steve Apr 02 '14 at 19:45

Why does MinGW 4.8 not treat ex.what() as virtual?

1 Answers1