问题描述
是否可以在没有接收器实例的情况下将信号连接到静态插槽?
Is it possible to connect a signal to static slot without receiver instance?
像这样:connect(&object, SIGNAL(some()), STATIC_SLOT(staticFooMember()));
Qt 文档中有一个带有 [static slot] 属性的 QApplication::closeAllWindows() 函数.文档中有一个使用它的示例:
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()));
是否允许执行相同的操作但不传递实例变量(例如，当一个类只有静态函数时)?
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();
};
<小时>
也许 Frank Osterfeld 是对的，在这种情况下最好使用单例模式，但我仍然很惊讶为什么这个功能还没有实现.
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 中这是可能的.
推荐答案
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，这是不允许的.相反，允许使用作为静态函数的插槽，但为了能够连接它，您需要一个实例.
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.
在他们的例子中，
connect(exitAct, SIGNAL(triggered()), qApp, SLOT(closeAllWindows()));
意味着比他们之前所说的
means than they previously called
QApplication* qApp = QApplication::instance();
连接对象的唯一接口是函数
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 )
你打算如何摆脱const QObject *receiver?
检查项目中的 moc 文件，它会自己说话.
Check the moc files in your project, it speaks by itself.
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