问题描述

我有函数 f;我想在启动 f 后抛出异常 1s.我无法修改 f().可以用c++实现吗?

I have got function f; I want to throw exception 1s after start f. I can't modify f(). It it possible to do it in c++?

try {
   f();
}
catch (TimeoutException& e) {
//timeout
}

推荐答案

您可以创建一个单独的线程来运行调用本身，并在主线程中等待一个条件变量，该条件变量将由执行调用的线程发出信号f 一旦它返回.诀窍是在 1s 超时的情况下等待条件变量，这样如果调用时间超过超时时间，您仍然会醒来，知道它，并能够抛出异常 - 全部在主线程中.这是代码(现场演示 此处):

You can create a separate thread to run the call itself, and wait on a condition variable back in your main thread which will be signalled by the thread doing the call to f once it returns. The trick is to wait on the condition variable with your 1s timeout, so that if the call takes longer than the timeout you will still wake up, know about it, and be able to throw the exception - all in the main thread. Here is the code (live demo here):

#include <iostream>
#include <chrono>
#include <thread>
#include <mutex>
#include <condition_variable>

using namespace std::chrono_literals;

int f()
{
    std::this_thread::sleep_for(10s); //change value here to less than 1 second to see Success
    return 1;
}

int f_wrapper()
{
    std::mutex m;
    std::condition_variable cv;
    int retValue;

    std::thread t([&cv, &retValue]() 
    {
        retValue = f();
        cv.notify_one();
    });

    t.detach();

    {
        std::unique_lock<std::mutex> l(m);
        if(cv.wait_for(l, 1s) == std::cv_status::timeout) 
            throw std::runtime_error("Timeout");
    }

    return retValue;    
}

int main()
{
    bool timedout = false;
    try {
        f_wrapper();
    }
    catch(std::runtime_error& e) {
        std::cout << e.what() << std::endl;
        timedout = true;
    }

    if(!timedout)
        std::cout << "Success" << std::endl;

    return 0;
}

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