Convert QQuickItem* to QObject*

  • I have succeed to create some signal an qml and some slot to deal with them.
    I have created an empty project with the new application and it works, but when i add the code to an older app it say
    cannot convert 'QQuickItem*' to 'QObject*' in initialization QObject *item = view2.rootObject()

    My function look like this:

    void fromQML::loadQuestion()
        view2.setSource (QUrl("qrc:/question.qml"));
        QObject *item = view2.rootObject(); //here is the error and doesn't work if i make cast
        loadProcess load;
        QObject::connect(item, SIGNAL(qmlYesSig()),&load, SLOT(forYes()));
        QObject::connect(item, SIGNAL(qmlNoSig()),&load, SLOT(forNo()));

    It the working application the code looks like this:

        QGuiApplication app(argc, argv);
        QQuickView view;
        view.setSource (QUrl("qrc:/main.qml"));
        QObject *item = view.rootObject();
        testClass myClass;
        QObject::connect(item, SIGNAL(qmlYesSig(QString)),
                         &myClass, SLOT(cppSlot(QString)));
        QObject::connect(item, SIGNAL(qmlNoSig(QString)),
                         &myClass, SLOT(cppSlot(QString)));

    From my point of view is the same staff so i don't understand why in first sample doesn't work and is the second does.

  • I have try again and if i make the connection in main it works but if i copy and paste the some code in a C++ slot it give me the above error.

  • Lifetime Qt Champion


    While that is surprising, why not use QQuickItem* item = view2.rootObject(); since it's what the method returns ?

    Note that your load object will be destroyed at the end of loadQuestion so it might not do anything.

Log in to reply