我想要达到的目标是，如果进程超过超时时间，程序将停止进程并返回超时错误。

我从 second most voted answer to this question 获得了一个超时函数。我遇到了一个错误，它说 std::result_of_t 在c++17中不受欢迎，所以我使用建议的替换 std::invoke_result_t 重新创建了它，并将其实现如下：

template <typename TF, typename TDuration, class... TArgs>
std::invote_result_t<TF&&, TArgs&&...> run_with_timeout(TF&& f, TDuration timeout, TArgs&&... args)
    using R = std::invoke_result_t<TF&&, TArgs&&...>;
    std::packaged_task<R(TArgs...)> task(f);
    auto future = task.get_future();
    std::thread thr(std::move(task), std::forward<TArgs>(args)...);
    if (future.wait_for(timeout) != std::future_status::timeout)
       thr.join();
       return future.get(); // this will propagate exception from f() if any
       thr.detach(); // we leave the thread still running
       throw std::runtime_error("Timeout");
}

我想使用它来检查类方法的函数是否超时。所以我尝试用一种类似于：

template <typename TF, typename TDuration, class... TArgs>
std::invote_result_t<TF&&, TArgs&&...> ClassOne::run_with_timeout(TF&& f, TDuration timeout, TArgs&&... args)
    using R = std::invoke_result_t<TF&&, TArgs&&...>;
    std::packaged_task<R(TArgs...)> task(f);
    auto future = task.get_future();
    std::thread thr(std::move(task), std::forward<TArgs>(args)...);
    if (future.wait_for(timeout) != std::future_status::timeout)
       thr.join();
       return future.get(); // this will propagate exception from f() if any
       thr.detach(); // we leave the thread still running
       throw std::runtime_error("Timeout");
// The function checked for timeout
int ClassOne::func(ClassTwo *param1, std::string name)
    // Some code here...
    // For mimicking function process:
    std::this_thread::sleep_for(10s);
    return 0;
// Function which calls the timed process with timeout function
int ClassOne::dummy(ClassTwo *param1, std::string name)
    int retVal = 0; // zero for no error, non-zero for error
        retVal = run_with_timeout(func, 20s, param1, name);