QTableWidget中修改数据如何发出SIGNAL

[weixin_38065454]

我修改了一下QT官网上的Chart的例子，在MainWindow.CPP中加入（粗体标出）：MainWindow::MainWindow(){    QMenu *fileMenu = new QMenu(tr("&File"), this);    QAction *openAction = fileMenu->addAction(tr("&Open..."));    openAction->setShortcuts(QKeySequence::Open);    QAction *saveAction = fileMenu->addAction(tr("&Save As..."));    saveAction->setShortcuts(QKeySequence::SaveAs);    QAction *quitAction = fileMenu->addAction(tr("E&xit"));    quitAction->setShortcuts(QKeySequence::Quit);    setupModel();    setupViews();    connect(openAction, SIGNAL(triggered()), this, SLOT(openFile()));    connect(saveAction, SIGNAL(triggered()), this, SLOT(saveFile()));    connect(quitAction, SIGNAL(triggered()), qApp, SLOT(quit()));   connect (model,SIGNAL(dataChanged(const QModelIndex & topLeft, const QModelIndex & bottomRight)),this,SLOT(display(const QModelIndex & topLeft, const QModelIndex & bottomRight)));    menuBar()->addMenu(fileMenu);    statusBar();    openFile(":/Charts/qtdata.cht");    setWindowTitle(tr("Chart"));    resize(870, 550);}我是想程序执行的时候如果修改了Table中的数据，则能够触发void  display(const QModelIndex & topLeft, const QModelIndex & bottomRight);这个我写的函数，但是发现程序执行的时候，无论我在表中怎么修改，都不会执行display()，请问这是为什么呢？期盼有高手帮我解答下这个问题。谢谢！