将QGraphicsview信号连接到UI插槽

Question

I am new to QT and C++ and i have legacy qt-c++ code here which i cant get to work. Probably its something about the lifetime of the calling objects but hey, but please tell me what i am missing.

In a QT .ui i have various Frames and Widgets containing Frames and Widgets containing a QVBoxLayout which we shall call "myLayout"

On click in the .ui file i use

myWidget = new mywidget(some params);
myLayout->addWidget(myWidget);

where myWidget is declared as mywidget *myWidget; in the header file

myWidget is a QWidget which internally adds a QVBoxLayout to itself and adds a QGraphicsView . Using the MouseReleaseEvent i emit a signal from the QWidget.

now when i try to connect the signal slot (which i do in cpp file from the ui)

connect(myWidget, SIGNAL(mySignal(QString)), this, SLOT(mySlot(QString)));

the signals never catch the slot. the slot is public, the signal isnt.

What did i do wrong? Can somebody help. Feel free to ask further questions since i dont really know whats important in c++ questions;)

edit: the signal gets emitted by QGraphicsObjects which themselfs connect to a slot in the QGraphicsView . This Slot is called and debuggable. at the end of the Routine an emit mySignal("..."); is called.

Answer 1

Maybe you forgot to add Q_OBJECT macro in your widget declaration. Qt documentation: http://doc.qt.io/qt-5/qobject.html#Q_OBJECT