我正在嘗試創建一個應該在後臺運行的新線程gpsthread,並存儲該值。使用Qthread-Qt5創建新線程
class gpsthread: public QThread{
Q_OBJECT
private:nrega_status_t status2;
public:
explicit gpsthread(QObject *parent = 0):QThread(parent) {
// QTimer *t = new QTimer(this);
// connect(t, SIGNAL(timeout()), this, SLOT(processgps()));
// t->start(10000);
}
void run(){
qDebug()<<"inside gps thread\n";
QTimer *t = new QTimer(this);
connect(t, SIGNAL(timeout()), this, SLOT(processgps()));
t->start(10000);
}
public slots:void processgps(){
int status2;
status2=gps_management();
}
};
我的主課是quickview。
int main(int argc, char *argv[])
{
QString file = "qml/main.qml";
QApplication app(argc, argv);
TranslationTest myObj;
QuickView view;
subthread object;
gpsthread obj;
gprsthread gprs;
view.rootContext()->setContextProperty("rootItem", (QObject *)&myObj);
obj.start();
//from subthread
QObject::connect(&object, SIGNAL(batterytoqml(QVariant,QVariant)),item, SLOT(frombattery(QVariant,QVariant)));
QObject::connect(&gprs, SIGNAL(gprstoqml(QVariant)),item, SLOT(fromgprs(QVariant)));
return app.exec();
}
我曾經嘗試這樣做,以及
class gpsthread: public QThread{
Q_OBJECT
private:nrega_status_t status2;
public:QTimer* t;
explicit gpsthread(QObject *parent = 0):QThread(parent) {
// QTimer *t = new QTimer(this);
// connect(t, SIGNAL(timeout()), this, SLOT(processgps()));
// t->start(10000);
}
void run(){
qDebug()<<"inside gps thread\n";
t = new QTimer(this);
connect(t, SIGNAL(timeout()), this, SLOT(processgps()));
t->start(10000);
exec();
}
public slots:void processgps(){
int status2;
status2=gps_management();
}
};
,但它給錯誤說
QObject: Cannot create children for a parent that is in a different thread
如果我在構造函數創建對象,然後也將會給相同的錯誤,因爲該對象將在主線程中創建。 如何解決這個問題?
感謝解釋這...這確實是使用線程的正確方法。 – Goddard