/EDIT: solved, see my comment in the 1st answer!/
I am currently building an application which only has a tray icon displayed, so it doesn't have any windows.
Well, in the tray icon I've included a QAction so as to close the application. The thing is, that I get seg fault when I call exit(0); from that function. This is some example code:
//I have a reason for setting it to be a QTimer, please don't even comment on this
class Boot_Timer : public QTimer {
Q_OBJECT
public:
explicit Boot_Timer(QObject *parent = 0) : QTimer(parent) {
}
public Q_SLOTS:
void set_up_command_line_tray(){
//Setting up the tray Icon.
QSystemTrayIcon *trayIcon_cmd = new QSystemTrayIcon(this);
trayIcon_cmd->setIcon(QIcon(":/icons/Pictures/myapp.png"));
trayIcon_cmd->setToolTip("My tray tooltipp");
QMenu *changer_menu = new QMenu;
QAction *Quit_action = new QAction(tr("&Quit"), this);
Quit_action->setIconVisibleInMenu(true);;
connect(Quit_action, SIGNAL(triggered()), this, SLOT(close_application()));
changer_menu->addAction(Quit_action);
trayIcon_cmd->setContextMenu(changer_menu);
trayIcon_cmd->show();
}
void close_application(){
//HERE I GET SEG FAULT
exit(0);
}
};
Boot_Timer boottimer;
#include "main.moc"
int main(int argc, char *argv[])
{
QApplication a(argc, argv);
//making some checks (code omitted)
...
boottimer.set_up_command_line_tray()
return app.exec();
}
So, the tray icon is shown normally and perfectly, but when I choose to Quit the application using the menu I've added to the tray icon, I get a seg fault. I guess that I cannot quit the application using exit(int state) outside main() function and its functions that don't have a parent...
What is the correct way to quit my application, then?
Thanks in advance for any answers!
Thanks, that didn't solve it. For some reason, the thing that solved it was to do the following: QSystemTrayIcon *trayIcon_cmd = new QSystemTrayIcon(0); instead of QSystemTrayIcon *trayIcon_cmd = new QSystemTrayIcon(this)