替换已弃用的 QtSignalMapper 类以在 Qt5 中转发信号

San*_*912 1 c++ qt qt5

我有这样的代码,它为 Qt 4 编写了一个 mdi 窗口:

class MdiWindow : public QMainWindow
{
    Q_OBJECT
public:
    MdiWindow( QWidget *parent = nullptr)

...
private:
    QWorkspace* workspace
    QSignalMapper* mapper
}


MdiWindow::MdiWindow( QWidget *parent ) : QMainWindow( parent )
{
  ...

  workspace = new QWorkspace;
  setCentralWidget( workspace );

  connect( workspace, SIGNAL(windowActivated(QWidget *)), this, SLOT(enableActions()));
  mapper = new QSignalMapper( this );
  connect( mapper, SIGNAL(mapped(QWidget*)), workspace, SLOT(setActiveWindow(QWidget*)) );

  ....
}
Run Code Online (Sandbox Code Playgroud)

根据 QT 文档QWorkspace应该替换为QMdiArea.

我这样做了并编写了第一个连接,如下所示:

connect(workspace, &QMdiArea::subWindowActivated,
        this, &MdiWindow::enableActions);
Run Code Online (Sandbox Code Playgroud)

但那QSignalMapper也被弃用了呢?

那么我该如何更新这一行:

mapper = new QSignalMapper( this );
connect( mapper, SIGNAL(mapped(QWidget*)), workspace, SLOT(setActiveWindow(QWidget*)) );
Run Code Online (Sandbox Code Playgroud)

我读到QSignalMapper可以用 lamdas 替换,但在这种情况下如何替换?如果我明白,则将mapper所有信号从这里转发到活动窗口workspace

Tup*_*oal 5

以前,您通常QSignalMapper::setMapping()要确保在调用时会向您发送所需的数据SLOT()。现在你可以将此逻辑封装在lambda中,所以如果你这样做了(就像在Qt示例中一样):

     for (int i = 0; i < texts.size(); ++i) {
         QPushButton *button = new QPushButton(texts[i]);
         connect(button, SIGNAL(clicked()), signalMapper, SLOT(map()));
         signalMapper->setMapping(button, texts[i]);
     }
     connect(signalMapper, SIGNAL(mapped(const QString &)),
             this, SIGNAL(clicked(const QString &)));
Run Code Online (Sandbox Code Playgroud)

你现在可以做(某种程度上):

     for (int i = 0; i < texts.size(); ++i) {
         QPushButton *button = new QPushButton(texts[i]);
         connect(button, &QPushButton::clicked, [=]() {
             emit clicked(texts[i]);
         });
     }
Run Code Online (Sandbox Code Playgroud)

如果setMapping()没有使用,那么它可能已经直接连接到了SLOT()