Question

我遇到了从另一个线程恢复 boost :: asio 协同程序的问题。以下是示例代码：

#include <iostream>
#include <thread>

#include <boost/asio.hpp>
#include <boost/asio/steady_timer.hpp>
#include <boost/asio/spawn.hpp>

using namespace std;
using namespace boost;

void foo(asio::steady_timer& timer, asio::yield_context yield)
{
    cout << "Enter foo" << endl;
    timer.expires_from_now(asio::steady_timer::clock_type::duration::max());
    timer.async_wait(yield);
    cout << "Leave foo" << endl;
}

void bar(asio::steady_timer& timer)
{
    cout << "Enter bar" << endl;
    sleep(1); // wait a little for asio::io_service::run to be executed
    timer.cancel();
    cout << "Leave bar" << endl;
}

int main()
{
    asio::io_service ioService;
    asio::steady_timer timer(ioService);

    asio::spawn(ioService, bind(foo, std::ref(timer), placeholders::_1));

    thread t(bar, std::ref(timer));

    ioService.run();
    t.join();

    return 0;
}

问题是 asio :: steady_timer 对象不是线程安全的，程序崩溃了。但是如果我尝试使用互斥锁来同步对它的访问，那么我就会出现死锁，因为 foo 的范围没有留下。

#include <iostream>
#include <thread>
#include <mutex>

#include <boost/asio.hpp>
#include <boost/asio/steady_timer.hpp>
#include <boost/asio/spawn.hpp>

using namespace std;
using namespace boost;

void foo(asio::steady_timer& timer, mutex& mtx, asio::yield_context yield)
{
    cout << "Enter foo" << endl;

    {
        lock_guard<mutex> lock(mtx);
        timer.expires_from_now(
            asio::steady_timer::clock_type::duration::max());
        timer.async_wait(yield);
    }

    cout << "Leave foo" << endl;
}

void bar(asio::steady_timer& timer, mutex& mtx)
{
    cout << "Enter bar" << endl;
    sleep(1); // wait a little for asio::io_service::run to be executed

    {
        lock_guard<mutex> lock(mtx);
        timer.cancel();
    }

    cout << "Leave bar" << endl;
}

int main()
{
    asio::io_service ioService;
    asio::steady_timer timer(ioService);
    mutex mtx;

    asio::spawn(ioService, bind(foo, std::ref(timer), std::ref(mtx),
        placeholders::_1));

    thread t(bar, std::ref(timer), std::ref(mtx));

    ioService.run();
    t.join();

    return 0;
}

如果我使用标准的完成处理程序而不是协同程序，则没有这样的问题。

#include <iostream>
#include <thread>
#include <mutex>

#include <boost/asio.hpp>
#include <boost/asio/steady_timer.hpp>

using namespace std;
using namespace boost;

void baz(system::error_code ec)
{
    cout << "Baz: " << ec.message() << endl;
}

void foo(asio::steady_timer& timer, mutex& mtx)
{
    cout << "Enter foo" << endl;
    {
        lock_guard<mutex> lock(mtx);
        timer.expires_from_now(
            asio::steady_timer::clock_type::duration::max());
        timer.async_wait(baz);
    }
    cout << "Leave foo" << endl;
}

void bar(asio::steady_timer& timer, mutex& mtx)
{
    cout << "Enter bar" << endl;
    sleep(1); // wait a little for asio::io_service::run to be executed
    {
        lock_guard<mutex> lock(mtx);
        timer.cancel();
    }
    cout << "Leave bar" << endl;
}

int main()
{
    asio::io_service ioService;
    asio::steady_timer timer(ioService);
    mutex mtx;

    foo(std::ref(timer), std::ref(mtx));

    thread t(bar, std::ref(timer), std::ref(mtx));

    ioService.run();
    t.join();

    return 0;
}

使用couroutines时，是否可能有与上一个示例类似的行为。

Answer 1

协程在strand的上下文中运行。在spawn()中，如果未明确提供，则会为协程创建新的strand。通过明确地将strand提供给spawn()，可以将工作发布到将与协程同步的strand。

此外，如sehe所述，如果协程在一个线程中运行，获取互斥锁，然后挂起，但在另一个线程中恢复并运行并释放锁，则可能会发生未定义的行为。为了避免这种情况，理想情况下，当协程暂停时，不应该持有锁。但是，如果有必要，必须保证协程在恢复时在同一个线程中运行，例如只运行单个线程中的io_service。

以下是基于原始示例的最小完整example，其中bar()发布工作到strand取消计时器，导致foo()协程恢复：< / p>

#include <iostream>
#include <thread>

#include <boost/asio.hpp>
#include <boost/asio/spawn.hpp>
#include <boost/asio/steady_timer.hpp>

void foo(boost::asio::steady_timer& timer, boost::asio::yield_context yield)
{
  std::cout << "Enter foo" << std::endl;

  timer.expires_from_now(
      boost::asio::steady_timer::clock_type::duration::max());
  boost::system::error_code error;
  timer.async_wait(yield[error]);
  std::cout << "foo error: " << error.message() << std::endl;

  std::cout << "Leave foo" << std::endl;
}

void bar(
  boost::asio::io_service::strand& strand,
  boost::asio::steady_timer& timer
)
{
  std::cout << "Enter bar" << std::endl;

  // Wait a little for asio::io_service::run to be executed
  std::this_thread::sleep_for(std::chrono::seconds(1));
  // Post timer cancellation into the strand.
  strand.post([&timer]()
    {
      timer.cancel();
    });

  std::cout << "Leave bar" << std::endl;
}

int main()
{
  boost::asio::io_service io_service;
  boost::asio::steady_timer timer(io_service);
  boost::asio::io_service::strand strand(io_service);

  // Use an explicit strand, rather than having the io_service create.
  boost::asio::spawn(strand, std::bind(&foo, 
      std::ref(timer), std::placeholders::_1));

  // Pass the same strand to the thread, so that the thread may post
  // handlers synchronized with the foo coroutine.
  std::thread t(&bar, std::ref(strand), std::ref(timer));

  io_service.run();
  t.join();
}

提供以下输出：

Enter foo
Enter bar
foo error: Operation canceled
Leave foo
Leave bar

正如this回答所述，当boost::asio::yield_context检测到异步操作失败时，例如取消操作时，它会将boost::system::error_code转换为system_error异常和抛出。上面的示例使用yield_context::operator[]允许yield_context在失败时填充提供的error_code，而不是投掷。

从另一个线程恢复asio协同程序

1 个答案: