我试图做一个简单的GUI与QT 4.6。我做了表示菜单栏,一个separete类:简单的菜单栏使用Qt4
MenuBar::MenuBar()
{
aboutAct = new QAction(tr("&About QT"), this);
aboutAct->setStatusTip(tr("Show the application's About box"));
connect(aboutAct, SIGNAL(triggered()), this, SLOT(about()));
quitAct = new QAction(tr("&Quit"),this);
quitAct->setStatusTip(tr("Exit to the program"));
//connect(quitAct, SIGNAL(triggered()), &QApp, SLOT(quit()));
menuFile = new QMenu("File");
menuFile->addAction(quitAct);
menuLinks = new QMenu("Links");
menuAbout = new QMenu("Info");
menuAbout->addAction(aboutAct);
addMenu(menuFile);
addMenu(menuLinks);
addMenu(menuAbout);
}
我不能可能是主要的应用程序的退出槽quitAct的信号连接,因为它是不是从MenuBar类可见..
//connect(quitAct, SIGNAL(triggered()), &QApp, SLOT(quit()));
我该怎么办?
所有的方法都这么 “脏”,而不是Qt的风格。而且你没有在他的代码中发现错误 – 2010-04-30 10:19:48