下面的程序永远不会到达处理程序()。我正在使用信号集安装我自己的信号处理程序。
void handler( const boost::system::error_code& error , int signal_number )
{
ROS_ERROR("inside signal handler");
exit(1);
}
int main( int argc , char** argv )
{
ros::init(argc, argv, "name", ros::init_options::NoSigintHandler);
boost::asio::io_service io_service;
// Construct a signal set registered for process termination.
boost::asio::signal_set signals(io_service, SIGINT );
// Start an asynchronous wait for one of the signals to occur.
signals.async_wait( handler );
boost::asio::spawn(io_service, {
while(1);
}
);
io_service.run();
return 0;
}有意思,当我用
signals.async_wait(
[&ioSservice](boost::system::error_code& code, int signalNo) {
ioService.stop();
});那它就不会终止。
发布于 2018-09-28 21:31:55
您只有一个线程为您的io_service服务,它忙于处理while(1);,因此不能运行信号处理程序。
io_service的作用类似于队列。当您对事物进行async_wait时,asio将安排将回调添加到队列中,以便通过相关的io_service运行。调用io_service::run时,调用线程将从io_service的队列中提取挂起的项并运行它们。
在本例中,当您调用io_service.run()时,队列中有任务:由spawn创建的任务,它运行着一个没完没了的while循环。由于循环从未结束,主线程永远无法完成该任务。稍后,当signal_set接收到SIGINT时,它会向队列中添加另一个作业以调用handler,但是它永远不会被运行,因为从队列中提取作业的唯一线程正忙于一个没完没了的while循环。
处理这一问题的方法是避免将长期运行的作业放在io_service的队列中和/或让多个线程为io_service服务。
void handler(const boost::system::error_code& error, int signal_number)
{
std::cout << "inside signal handler\n";
exit(1);
}
int main(int argc, char** argv)
{
boost::asio::io_service io_service;
// You can use a work object to avoid having the io_service
// stop when its job queue empties.
boost::asio::io_service::work work(io_service);
boost::asio::signal_set signals(io_service, SIGINT);
signals.async_wait(handler);
// Now that there's a work object keeping this io_service running
// this call isn't needed at all. It's just here to demonstrate
// that it works
boost::asio::spawn(io_service, []{
while(1);
}
);
// Start a second thread to run io_service jobs
std::thread t([&io_service]{ io_service.run(); });
// Also handle io_service jobs on this thread
io_service.run();
return 0;
}https://stackoverflow.com/questions/52562406
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