Creating a Qt Quick Controls project in Qt Creator 3.0 produces this template code:

#include "qtquick2controlsapplicationviewer.h"

int main(int argc, char *argv[])

{ Application app(argc, argv); QtQuick2ControlsApplicationViewer viewer; viewer.setMainQmlFile(QStringLiteral("qml/QtQuickControlsTest/main.qml")); viewer.show(); return app.exec(); }

However, there is no way to use the QML file as a resource with QtQuick2ControlsApplicationViewer. Simply placing the QML file into a resource file and modifying the generated code to use it leads to an error indicating "Your root item has to be a Window."

A workaround is to modify the generated code to use QQmlApplicationEngine instead.