JuniorTheoryCommon

What does the program print?

C++

#include <iostream>

struct Error {
    virtual int code() const { return 1; }
};
struct NotFound : public Error {
    int code() const override { return 404; }
};
int main() {
    try { throw NotFound{}; } catch (Error e) { std::cout << e.code() << ' '; }
    try { throw NotFound{}; } catch (const Error& e) { std::cout << e.code(); }
}

1 404

Expecting 404 404: catch (Error e) copies only the Error part of the thrown NotFound, so code() runs on a plain Error.
Expecting 1 1: const Error& e refers to the thrown NotFound itself, and a virtual call through it runs NotFound::code.

Why does GCC warn about catch (Error e) when it builds with -Wall?
Why do you throw the exception object itself rather than a pointer to it?

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JuniorTheoryCommon

What does the program print?

C++

#include <iostream>

struct Error {
    virtual int code() const { return 1; }
};
struct NotFound : public Error {
    int code() const override { return 404; }
};
int main() {
    try { throw NotFound{}; } catch (Error e) { std::cout << e.code() << ' '; }
    try { throw NotFound{}; } catch (const Error& e) { std::cout << e.code(); }
}

1 404

Expecting 404 404: catch (Error e) copies only the Error part of the thrown NotFound, so code() runs on a plain Error.
Expecting 1 1: const Error& e refers to the thrown NotFound itself, and a virtual call through it runs NotFound::code.

Why does GCC warn about catch (Error e) when it builds with -Wall?
Why do you throw the exception object itself rather than a pointer to it?

Back to questions Open topic in roadmap →Answer it yourself — Quiz