問題描述
是否可以在沒有接收器實例的情況下將信號連接到靜態插槽?
Is it possible to connect a signal to static slot without receiver instance?
像這樣:connect(&object, SIGNAL(some()), STATIC_SLOT(staticFooMember()));
Qt 文檔中有一個帶有 [static slot] 屬性的 QApplication::closeAllWindows() 函數.文檔中有一個使用它的示例:
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()));
是否允許執行相同的操作但不傳遞實例變量(例如，當一個類只有靜態函數時)?
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();
};
<小時>
也許 Frank Osterfeld 是對的，在這種情況下最好使用單例模式，但我仍然很驚訝為什么這個功能還沒有實現.
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.
更新:
在 Qt 5 中這是可能的.
推薦答案
QT5 更新:是的，你可以
static void someFunction() {
qDebug() << "pressed";
}
// ... somewhere else
QObject::connect(button, &QPushButton::clicked, someFunction);
在 QT4 中你不能:
不，這是不允許的.相反，允許使用作為靜態函數的插槽，但為了能夠連接它，您需要一個實例.
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.
在他們的例子中，
connect(exitAct, SIGNAL(triggered()), qApp, SLOT(closeAllWindows()));
意味著比他們之前所說的
means than they previously called
QApplication* qApp = QApplication::instance();
連接對象的唯一接口是函數
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 )
你打算如何擺脫const QObject *receiver?
檢查項目中的 moc 文件，它會自己說話.
Check the moc files in your project, it speaks by itself.
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