JuniorTheoryCommon

What happens when this program runs?

C++

#include <iostream>
#include <stdexcept>

struct Logger {
    ~Logger() { throw std::runtime_error("flush failed"); }
};

int main() {
    try { Logger log; }
    catch (const std::exception&) { std::cout << "caught"; }
}

The program calls std::terminate: the destructor is noexcept

Expecting caught: a destructor is noexcept by default, so an exception leaving it calls std::terminate, and no handler runs.
Expecting a compile error: a throw in a destructor builds, and GCC and Clang only warn about it.

How can Logger report a failed flush without throwing from its destructor?
What changes if the destructor catches its own exception inside a try block?

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JuniorTheoryCommon

What happens when this program runs?

C++

#include <iostream>
#include <stdexcept>

struct Logger {
    ~Logger() { throw std::runtime_error("flush failed"); }
};

int main() {
    try { Logger log; }
    catch (const std::exception&) { std::cout << "caught"; }
}

The program calls std::terminate: the destructor is noexcept

Expecting caught: a destructor is noexcept by default, so an exception leaving it calls std::terminate, and no handler runs.
Expecting a compile error: a throw in a destructor builds, and GCC and Clang only warn about it.

How can Logger report a failed flush without throwing from its destructor?
What changes if the destructor catches its own exception inside a try block?

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