Question

此代码旨在使用Boost.Asio接收UDP多播消息。当在接收者的构造函数中进行第二次set_option（）调用（加入多播组）时，下面的代码抛出Boost system_error异常。投诉是“无效论证”。这似乎与构造函数发生在IO :: doIO（）中定义的lambda内部的事实有关，因为使用具有相同功能的std :: thread的成员（IO :: threadFunc（））会导致预期行为（没有抛出异常）。

为什么会这样，我该如何修复它以便我可以使用lambda？

//g++ -std=c++11 doesntWork.cc -lboost_system -lpthread

#include <iostream>
#include <thread>
#include <boost/asio.hpp>
#include <boost/bind.hpp>

class IO
{
public:
   class receiver
   {
   public:
      receiver(
         boost::asio::io_service         &io_service,
         const boost::asio::ip::address  &multicast_address,
         const unsigned short             portNumber) : _socket(io_service)
      {
         const boost::asio::ip::udp::endpoint listen_endpoint(
            boost::asio::ip::address::from_string("0.0.0.0"), portNumber);

         _socket.open(listen_endpoint.protocol());
         _socket.set_option(boost::asio::ip::udp::socket::reuse_address(true));
         _socket.bind(listen_endpoint);
std::cerr << " About to set option join_group" << std::endl;
         _socket.set_option(boost::asio::ip::multicast::join_group(
            multicast_address));

         _socket.async_receive_from(
            boost::asio::buffer(_data),
            _sender_endpoint,
            boost::bind(&receiver::handle_receive_from, this,
               boost::asio::placeholders::error,
               boost::asio::placeholders::bytes_transferred));
      }

   private:
      void handle_receive_from(
         const boost::system::error_code &error,
         const size_t                     bytes_recvd)
      {
         if (!error)
         {
            for(const auto &c : _data)
               std::cout << c;
            std::cout << std::endl;
         }
      }

   private:
      boost::asio::ip::udp::socket   _socket;
      boost::asio::ip::udp::endpoint _sender_endpoint;
      std::vector<unsigned char> _data;
   }; // receiver class


   void doIO()
   {
      const boost::asio::ip::address multicast_address =
         boost::asio::ip::address::from_string("235.0.0.1");

      const unsigned short portNumber = 9999;

//       _io_service_thread = std::thread(
//          &IO::threadFunc, this, multicast_address, portNumber);

      _io_service_thread = std::thread([&, this]{
         try {
            // Construct an asynchronous receiver
            receiver r(_io_service, multicast_address, portNumber);

            // Now run the IO service
            _io_service.run();
         }
         catch(const boost::system::system_error &e)
         {
            std::cerr << e.what() << std::endl;
            throw e; //std::terminate()
         }
      });
   }

   void threadFunc(
      const boost::asio::ip::address &multicast_address,
      const unsigned short portNumber)
   {
      try {
         // Construct an asynchronous receiver
         receiver r(_io_service, multicast_address, portNumber);

         // Now run the IO service
         _io_service.run();
      }
      catch(const boost::system::system_error &e)
      {
         std::cerr << e.what() << std::endl;
         throw e; //std::terminate()
      }
   }

private:
   boost::asio::io_service _io_service;
   std::thread             _io_service_thread;
}; // IO class

int main()
{
   IO io;
   io.doIO();

   std::cout << "IO Service is running" << std::endl;

   sleep(9999);
}

Answer 1

存在一种竞争条件，可能导致访问悬空引用，从而调用未定义的行为。 lambda捕获列表通过引用捕获自动变量multicast_address和portNumber。但是，这些对象的生命周期可能会在_io_service_thread中使用之前结束：

void doIO()
{
  const boost::asio::ip::address multicast_address = /* ... */;
  const unsigned short portNumber = /* ... */;

  _io_service_thread = std::thread([&, this] {
    // multicast_address and portNumber's lifetime may have already ended.
    receiver r(_io_service, multicast_address, portNumber);
    // ...
  });
} // multicast_address and portNumber are destroyed.

要解决此问题，请考虑按值捕获，以便线程对生命周期一直有效的副本进行操作，直到线程结束。变化：

std::thread([&, this] { /* ... */ }

为：

std::thread([=] { /* ... */ }

当使用函数及其所有参数构造std::thread时，此问题不会出现，因为std::thread构造函数会将所有提供的参数复制/移动到线程可访问的存储中。

另外，如果_io_service_thread对象仍然可以在std::terminate()的析构函数中连接，请注意IO对象的销毁将调用_io_service_thread。要避免此行为，请考虑从主线程明确加入{{1}}。

set_option：在lambda中设置选项boost :: asio :: ip :: multicast :: join_group时的参数无效

1 个答案: