JuniorTheoryCommon

What happens when this program runs?

C++

#include <iostream>
#include <stdexcept>

void check_age(int age) noexcept {
    if (age < 0) {
        throw std::invalid_argument("negative age");
    }
}
int main() {
    try { check_age(-1); }
    catch (const std::exception&) { std::cout << "caught"; }
}

The program calls std::terminate: check_age is noexcept

Expecting caught: an exception that would leave a noexcept function calls std::terminate instead, so no outer handler sees it.
Expecting a compile error: the compiler does not enforce noexcept; GCC and Clang only warn, and the program builds.

What would the program print if check_age caught its own exception inside a try block?
Why is a function that may fail usually not marked noexcept?

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JuniorTheoryCommon

What happens when this program runs?

C++

#include <iostream>
#include <stdexcept>

void check_age(int age) noexcept {
    if (age < 0) {
        throw std::invalid_argument("negative age");
    }
}
int main() {
    try { check_age(-1); }
    catch (const std::exception&) { std::cout << "caught"; }
}

The program calls std::terminate: check_age is noexcept

Expecting caught: an exception that would leave a noexcept function calls std::terminate instead, so no outer handler sees it.
Expecting a compile error: the compiler does not enforce noexcept; GCC and Clang only warn, and the program builds.

What would the program print if check_age caught its own exception inside a try block?
Why is a function that may fail usually not marked noexcept?

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