I am using QProcess::start to launch Qt Assistant with my custom help project file. It works fine until i load project(not help project file) to my programm. Programm generates images from specific data using custom library. Even when all processes ends and i see generated images and nothing else happens, when i trying to launch Qt Assistant, my programm hangs at QProcess:start function when trying to start process. The code is:
show() function(public):
if (!run())
return false;
QByteArray ba("setSource ");
ba.append("qthelp://insyn_help/doc/");
proc->write(ba + page.toLocal8Bit() + '\n');
return true;
run() function(private):
if (!proc)
proc = new QProcess();
if (proc->state() == QProcess::Running)
return true;
QString app = QString(QT_BIN_DIR) + QDir::separator() + QString("assistant");
QString path = QString(PREFIX) + QString(HELP_INSTALL_PATH) + QString("/help_project.qhc");
QStringList args;
args << QLatin1String("-collectionFile")
<< QLatin1String(path.toLatin1())
<< QLatin1String("-enableRemoteControl");
QFileInfo help_project(path);
if (help_project.exists()) {
proc->start(app,args);
if (!proc->waitForStarted()) {
m_exitCode = 1;
emit closed();
return false;
}
}
This code is a part of AssistantLauncher class which was registered using qmlRegisterType and added to main.qml as a member of application window. My programm doesn't touch it anywhere (except calling a method show()). It is separate object (except it is a part of appWindow). The question is why does the process can not start only after my programm did some work? And why QProcess::start even dont have timeout.
UPD: I moved proc->start(app,args); to the child process, which i getting by using fork() and now my programm hangs on pid_t child = fork(). So the problem is that new process can not be created.
The answer is to do not use
fork()because it is dangerous in big projects. More at http://www.evanjones.ca/fork-is-dangerous.html .posix_spawnalso hangs my project. Now i decided tofork()new process at the beginning and send commands to it through the pipe.