是否保证未捕获的异常消息

Question

就是下面的代码

#include <stdexcept>

int main() {
    throw std::runtime_error("foobar");
}

保证产生以下结果？

terminate called after throwing an instance of 'std::runtime_error'
  what():  foobar
fish: Job 1, './a.out' terminated by signal SIGABRT (Abort)

我可以在每个编译器上依赖这个准确的输出吗？我可以依赖 what 方法将被调用并打印错误消息吗？

Answer 1

不保证，不确定是否有消息。来自 cppreference:

If an exception is thrown and not caught, including exceptions that escape the initial function of std::thread, the main function, and the constructor or destructor of any static or thread-local objects, then std::terminate is called. It is implementation-defined whether any stack unwinding takes place for uncaught exceptions.

这里相关的案例是“逃避 [...] 主要功能的异常”。不过，对 std::terminate 的调用是有保证的。如果你愿意，你可以安装一个 std::terminate_handler 来打印自定义消息。

是否保证未捕获的异常消息

Is uncaught exception message guaranteed

c++

exception