c++ - 线程如何在完成时发出信号？

Question

#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;
}

为什么上面程序的输出是:

Begin..
Cool, it's running.
dostuff beginning 4
dostuff is done doing stuff
still running? why

dostuff 如何在完成时正确标记？我不想坐等它，我只想在它完成时收到通知。

Answer 1

这个例子中的问题是 dostuff 有两个实例，所以 operator() 中设置为 false 的版本与 main 中的不同。

来自thread management documentation :

A new thread is launched by passing an object of a callable type that can be invoked with no parameters to the constructor. The object is then copied into internal storage, and invoked on the newly-created thread of execution. If the object must not (or cannot) be copied, then boost::ref can be used to pass in a reference to the function object. In this case, the user of Boost.Thread must ensure that the referred-to object outlives the newly-created thread of execution.

如果你不想复制对象，使用boost::ref:

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