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我一直在编写一个 linux 守护程序，它在 TCP/IP 上侦听请求并在接收到该请求时启动应用程序。我的问题是当我从命令提示符或 IDE (eclipse 3.7) 运行此守护程序时，一切正常并且我的可执行文件启动。但是当我使用
sudo service <myservicename> start 它时，它将在套接字上接收请求，但它不会启动该可执行文件。

这是我用于守护进程的标准代码 /// Linux 守护进程相关的东西

/// Create the lock file as the current user 
int lfp = open( "/var/lock/subsys/LauncherService", O_RDWR|O_CREAT, 0640);
if ( lfp < 0 ) 
{
    LOG_ERROR ("Unable to open lockfile");
    LOG_ERROR ( strerror(errno) );
}

/// All
/// Our process ID and Session ID
pid_t pid, sid;

/// Fork off the parent process
pid = fork();
if (pid < 0) {
       exit(EXIT_FAILURE);
}

/// If we got a good PID, then
///  we can exit the parent process.
if (pid > 0)
{
    exit(EXIT_SUCCESS);
}

/// Change the file mode mask
umask(0);

/// Create a new SID for the child process
sid = setsid();
if (sid < 0)
{
    LOG_ERROR ("Error Setting sid");
    exit(EXIT_FAILURE);
}
LOG_INFO ("sid set");

/// Change the current working directory
if ( (chdir("/usr/local/<mylocaldir>/bin")) < 0 )
{
       LOG_ERROR ("Error changing Directory");
       exit(EXIT_FAILURE);
}
LOG_INFO ("chdir successful");

/* Close out the standard file descriptors */
close(STDIN_FILENO);
close(STDOUT_FILENO);
close(STDERR_FILENO);

这是我启动需要启动的二进制文件的功能。这个函数在分叉的子进程中被调用。

std::string configFile = "/usr/local/<mydir>/config/Settings.config";
std::string binary = "<binaryname>";
std::string path ="/usr/local/<mydir>/bin/<binaryname>";

//char *argv[] = { "<binaryname>", "/usr/local/<mydir>/config/Settings.config", (char *)0 };

LOG_INFO("Calling Process" );
if ( execlp( path.c_str(), binary.c_str(), configFile.c_str(), (char *)0 ) == -1 )
//if ( execv("/usr/local/<mydir>/bin/<binaryname>", argv) == -1 )
//if ( execvp("/usr/local/<mydir>/bin/<binaryname>", argv) == -1 )
{
    LOG_ERROR("System call failed !!")
    std::string errString = strerror(errno);
    LOG_ERROR (errString );
}
else
{        
    LOG_INFO("System call successful");
}
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1 回答 1

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So after discussion with Casey I investigated more into my called program and I found that my program indeed is getting called. Also I found out that environment variables are not the issue child process is taking environment from parent itself. I am creating QApplication (qt gui application) in my main program. Its some issue with that and linux system daemon. I will try to figure that out and will ask separate question if needed.

Edit: Final Solution It was a qt GUI application which was not able to connect to XServer. I had to changes suggested by Casey and given in this post Cannot connect to X server :0.0 with a Qt application

after that it started launching.

于 2013-06-07T20:11:57.340 回答