1
我相信新的QT快速通信,我試圖寫一個代碼來調用上QML信號錯誤鑄造rootobject()在QT main.cpp中用於與QML
//main.cpp
#include "qtquick1applicationviewer.h"
#include "QApplication"
#include"authenticate.h"
int main(int argc, char *argv[])
{
QApplication app(argc, argv);
QDeclarativeView view(QUrl::fromLocalFile("MyItem.qml"));
QObject *item = view.rootObject();
Authenticate myClass;
QObject::connect(item, SIGNAL(qmlSignal(QString)),
&myClass, SLOT(cppSlot(QString)));
view.show();
return app.exec();
}
誤差C++時隙是: 主的.cpp:15:錯誤:不能在初始化 QObject的轉換 'QGraphicsObject *' 到 '的QObject *' *項= viewer.rootObject();
//main.qml
import QtQuick 1.0
Item {
id: item
width: 100; height: 100
signal qmlSignal(String msg)
MouseArea {
anchors.fill: parent
onClicked: item.qmlSignal("Hello from QML")
}
}
^