[SOLVED]qobject_cast<t> confusion

[manny]

hello,

I was using an example out of a book callled "programming a gui with qt4 second edition" (running qt5). bassically in an example in which we were making a spreadsheet, the author used the following algorithm to open a file in the recent files section.

void MainWindow::openRecentFiles()
{
if(okToContinue())
{
QAction *action = qobject_cast<QAction *>(sender());

       if(action)
       {
           //loadFile takes in the files name and loads the file
           loadFile(action->data().toString)
       }
    
    }

}

I know that sender() will get the value of the sender of the signal an return a pointer to the object that called it. qobject_cast will then use it's argument (sender()) to cast it to the specified type(QAction). however can't i just do the following:

    QAction *action = sender();

any help is greatly appreciated

[Leonardo]

qobject_cast is more like a safe way of casting. When the conversion is not possible, it returns NULL. Doing like you said, an impossible conversion would raise a runtime error.

[SGaist]

Hi,

No because sender() returns a QObject * and not the type of object that emitted the signal. You could also be calling that slot from different places like a QPushButton or even directly from the code thus adding more complexity on the handling inside your slot.

Hope it helps

[manny]

Thanks for the replies, really helps :)