4

我有一个简单的类，当我的程序获得和失去焦点时停止和启动一个计时器，但它给出的错误是 QObject 是 MyApp 在每个信号槽连接上的模糊基础。以下是相关代码：

class MyApp : public QApplication, public QObject
{
    Q_OBJECT
    ...
}

这是我的（凌乱的）Main.cpp：

    #include <QtGui/QApplication>
    #include "qmlapplicationviewer.h"
    #include <QObject>
    #include <QGraphicsObject>
    #include <QTimer>
    #include <QVariant>
    #include "timecontrol.h"
    #include "scorecontrol.h"
    #include "Retry.h"
    #include <QEvent>
    #include "myapp.h"

    int main(int argc, char *argv[])
    {
        MyApp app(argc, argv);

        QmlApplicationViewer viewer;
        viewer.setOrientation(QmlApplicationViewer::ScreenOrientationLockLandscape);
        viewer.setMainQmlFile(QLatin1String("qml/Raker/main.qml"));
        viewer.showExpanded();

        QObject *rootObject = viewer.rootObject();

        QTimer *timmer = new QTimer;
        timmer->setInterval(1000);

        TimeControl *timcon = new TimeControl;

        scorecontrol *scorer = new scorecontrol;

        Retry *probeer = new Retry;

        QObject::connect(timmer, SIGNAL(timeout()), timcon, SLOT(updateTime()));
        QObject::connect(timcon, SIGNAL(setTime(QVariant)), rootObject, SLOT(setTime(QVariant)));
        QObject::connect(rootObject, SIGNAL(blockClicked(int, int)), scorer, SLOT(checkRight(int, int)));
        QObject::connect(scorer, SIGNAL(setScore(QVariant)), rootObject, SLOT(setScore(QVariant)));
        QObject::connect(scorer, SIGNAL(setState(QVariant)), rootObject, SLOT(setState(QVariant)));

        QObject::connect(rootObject, SIGNAL(start()), probeer, SLOT(Reetry()));
        QObject::connect(probeer, SIGNAL(start()), timmer, SLOT(start()));
        QObject::connect(probeer, SIGNAL(stop()), timmer, SLOT(stop()));
        QObject::connect(probeer, SIGNAL(start(int)), scorer, SLOT(randomNum(int)));
        QObject::connect(probeer, SIGNAL(sReset()), timcon, SLOT(reset()));
        QObject::connect(probeer, SIGNAL(tReset()), scorer, SLOT(reset()));
        QObject::connect(timcon, SIGNAL(timeOut()), scorer, SLOT(reset()));

        QObject::connect(timcon, SIGNAL(setState(QVariant)), rootObject, SLOT(setState(QVariant)));
        QObject::connect(timcon, SIGNAL(changeFinal()), scorer, SLOT(changeFinal()));
        QObject::connect(scorer, SIGNAL(setFinal(QVariant)), rootObject, SLOT(setFinal(QVariant)));

        QObject::connect(&app, SIGNAL(focusL()), probeer, SLOT(focusL()));
        QObject::connect(&app, SIGNAL(focusG()), probeer, SLOT(focusG()));

        return app.exec();
    }

我的应用程序.cpp：

    #include "myapp.h"
    #include <QDebug>
    #include <QObject>

    MyApp::MyApp(int argc, char **argv): QApplication(argc, argv)
    {
        installEventFilter(this);
    }

    bool MyApp::eventFilter(QObject *object, QEvent *event)
    {
        if (event->type() == QEvent::ApplicationDeactivate)
        {
            qDebug() << "Focus lost";
            focusL();
        }
        if (event->type() == QEvent::ApplicationActivate)
        {
            qDebug() << "Focus gained";
            focusG();
        }

        return false;
    }
4

2 回答 2

10

在您当前的示例中，您创建了一个拆分继承方案，其中您的对象以 a 的双重实例结束QObject......有一个QObject用于 的基础QApplication，另一个用于实际MyApp类。这会产生歧义，因为访问继承的QObject方法或数据成员将无法确切知道要访问哪个继承的基础对象。

现在，您的继承图如下所示（注意您的MyApp对象继承的两个 QObject 实例）：

 | QObject |         | QObject |
       \                 /
        \         | QApplication |
         \             /
          \           /
         |    MyApp    |

你应该保持你的继承图是线性的，而不是有一个拆分继承方案，这意味着有一个只包含一个基类实例的派生类。所以你想要这样的东西：

    QObject
       |
       |
  QApplication
       |
       |
     MyApp
于 2011-10-26T13:45:02.747 回答
4

QApplication已经是QObject，所以你应该简单地写：

class MyApp : public QApplication
{
    Q_OBJECT
...

}
于 2011-10-26T13:36:38.657 回答