gtkmm的g_signal_emit或g_signal_emit_by_name版本是什么？

我正在程序后台运行一个超时函数，并试图从Gtk:：Button发出删除事件信号，这是我的构造函数中的代码片段：

// Glib::SignalProxy1<bool,GdkEventAny*> m_deleteSlot;
// m_deleteSlot =
signal_delete_event().connect (sigc::mem_fun (*this, &AlarmUI::my_delete_event));
m_timeout_connection = Glib::signal_timeout().connect_seconds(sigc::mem_fun(*this, &AlarmUI::cb_my_tick), 1);`

现在，方法：

bool AlarmUI::my_delete_event (GdkEventAny *event) {
if (m_timeout_connection.connected ()) {
    // show messagebox here
    return true;
} else {
    // bye bye
    return false;
}
}

现在，当用户单击退出按钮时，我想发出删除事件信号。问题：如何在gtkmmm中像在C g_signal_mit或g_signal_mit_by_name中那样发射信号？

void AlarmUI::on_button_quit () {
// m_deleteSlot.emit (); ???
}

更新1:

Glib::RefPtr<Gtk::Application> app = Gtk::Application::create (argc, argv, PACKAGE);
Glib::RefPtr<Gtk::Builder> refBuilder = Gtk::Builder::create ();
try { 
    refBuilder->add_from_file (UI_PATH);
}
catch (const Glib::FileError& ex) {
    std::cout << "FileError: " << ex.what() << std::endl;
    return 1;
}
catch (const Gtk::BuilderError& ex) {
    std::cout << "BuilderError: " << ex.what() << std::endl;
    return 1;
}
catch(const Glib::MarkupError& ex)
{
    std::cout << "MarkupError: " << ex.what() << std::endl;
    return 1;
}
AlarmUI *ui = 0;
refBuilder->get_widget_derived ("window1", ui);
if (ui) {
    ui->show_all ();
    app->run (); // The window doesn't show
}
delete ui;