Qt:发出信号时不调用信号处理程序

Question

我正在做这样的事情:

MyClass::MyClass(QWidget *parent) : QLabel(parent)
{
    this->MyMenu = new QMenu();

    QAction* act_del = new QAction(MyMenu);
    act_delete->setText("MyAction");

    MyMenu->addAction(act_del);

    QObject::connect(act_del,SIGNAL(triggered()),this,SLOT(MySlot()));
}

void MyClass::MySlot()
{
    //Not called
}

关于何时触发SIGNAL的任何建议都不会被调用.这是显示菜单的位置:

void MyClass::contextMenuEvent(QContextMenuEvent *ev)
{
    QPoint globalPos = this->mapToGlobal(ev->pos());

    QAction* selectedItem = MyMenu->exec(globalPos);
    if (selectedItem)
    {

    }
    else
    {
        // nothing was chosen
    }
}

为什么没有调用SLOT的任何建议？

Answer 1

MyClass需要包含Q_OBJECT宏以使信号槽连接起作用.