Is it possible to connect a signal to a static slot without a receiver instance?

bartolo-otrit picture bartolo-otrit · Feb 24, 2012 · Viewed 21.1k times · Source

Is it possible to connect a signal to static slot without receiver instance?

Like this: connect(&object, SIGNAL(some()), STATIC_SLOT(staticFooMember()));

There is a QApplication::closeAllWindows() function with [static slot] attribute in Qt documentation. And there is an example of using it from the documentation:

exitAct = new QAction(tr("E&xit"), this);
exitAct->setShortcuts(QKeySequence::Quit);
exitAct->setStatusTip(tr("Exit the application"));
connect(exitAct, SIGNAL(triggered()), qApp, SLOT(closeAllWindows()));

Is it allowed to do the same action but without passing an instance variable (e.g. when a class has only static functions)?

class Some : public QObject {
    Q_OBJECT
public slots:
    static void foo();
private:
    Some();
};

Maybe Frank Osterfeld is right and it is better to use singleton pattern in this case but I am still surprised why this feature has not been implemented yet.

Update:

In Qt 5 it is possible.

Answer

UmNyobe picture UmNyobe · Feb 24, 2012

Update for QT5: Yes you can

static void someFunction() {
    qDebug() << "pressed";
}
// ... somewhere else
QObject::connect(button, &QPushButton::clicked, someFunction);

In QT4 you can't:

No it is not allowed. Rather, it is allowed to use a slot which is a static function, but to be able to connect it you need an instance.

In their example,

connect(exitAct, SIGNAL(triggered()), qApp, SLOT(closeAllWindows()));

means than they previously called

QApplication* qApp = QApplication::instance();

Edit:

The only interface for connecting object is the function

bool QObject::connect ( const QObject * sender, const QMetaMethod & signal, const QObject * receiver, const QMetaMethod & method, Qt::ConnectionType type = Qt::AutoConnection )

How are you going to get rid of const QObject * receiver?

Check the moc files in your project, it speaks by itself.