c++ - 线程如何在完成时发出信号？

#include <iostream>
#include <boost/thread.hpp>
using std::endl; using std::cout;
using namespace boost;


mutex running_mutex;

struct dostuff
{
    volatile bool running;
    dostuff() : running(true) {}
    void operator()(int x)
    {
        cout << "dostuff beginning " << x << endl;
        this_thread::sleep(posix_time::seconds(2));
        cout << "dostuff is done doing stuff" << endl;
        mutex::scoped_lock running_lock(running_mutex);
        running = false;
    }
};

bool is_running(dostuff& doer)
{
    mutex::scoped_lock running_lock(running_mutex);
    return doer.running;
}

int main()
{
    cout << "Begin.." << endl;
    dostuff doer;
    thread t(doer, 4);

    if (is_running(doer)) cout << "Cool, it's running.\n";

    this_thread::sleep(posix_time::seconds(3));

    if (!is_running(doer)) cout << "Cool, it's done now.\n";
    else cout << "still running? why\n"; // This happens! :(

    return 0;
}

为什么上面的程序输出:



dostuff完成时如何正确标记？我不想坐在那里等着它，我只想在完成时收到通知。

最佳答案

此示例中的问题是存在dostuff的两个实例，因此operator()中设置为false的版本与main中的版本不同。

从thread management documentation:



如果您不想复制对象，请使用boost::ref:

thread t(boost::ref(doer), 4);