QObject连接“不能将参数3从继承QObject中的类转换为QObject*

Question

我正在使用MVC模式，并试图将我的视图类的信号与继承QObject的控制器类连接起来。

class View : public QWidget
{
    Q_OBJECT

private:

   Controller* controller;

   QPushButton* startButton;

   void addControls(QVBoxLayout* mainLayout);

public:
   explicit View(QWidget *parent = nullptr);

   void setController(Controller* c);
};

#endif // VIEW_H

以下是这些方法

void View::addControls(QVBoxLayout *mainLayout)
{
    //I'm adding the button
}

View::View(QWidget *parent) : QWidget(parent)
{
//Layout
}

void View::setController(Controller *c){
    controller = c;

    connect(startButton, SIGNAL(clicked()), controller, SLOT(begin()));
//ERROR controller is a Controller* and it can't be converted it to const QObject*
}

这是Controller类

class Controller : public QObject
{
    Q_OBJECT
private:
    QTimer* timer;

    View* view;
    Model* model;

public:
    explicit Controller(QObject *parent = nullptr);
    ~Controller();

    void setModel(Model* m);
    void setView(View* v);

public slots:
    void begin() const;

};

#endif // CONTROLLER_H

以及方法

Controller::Controller(QObject *parent):
    QObject(parent), timer(new QTimer)
{
    connect(timer, SIGNAL(timeout()), this, SLOT(next()));
}

Controller::~Controller() { delete timer; }

void Controller::setModel(Model* m) { model = m; }

void Controller::setView(View* v) { view = v; }

void Controller::begin() const {
    timer->start(200);

}

为了更好地衡量，这里是我设置每个组件的主要位置

int main(int argc, char *argv[])
{
    QApplication a(argc, argv);
    View w;
    Controller c;
    Model m;

    c.setModel(&m);
    c.setView(&w);
    w.setController(&c);
    w.show();
    return a.exec();
}

我已经试过了我能想到的一切，都做不到.

Answer 1

停止使用糟糕的基于文本的插槽连接接口。使用具有良好编译错误的现代版本：

QObject::connect(startButton, &QPushButton::clicked, controller, &Controller::begin);