QScopedPointer object, signal/slot usage, I am confused.

Pl45m4

Re: Hybrid Qt GUI/Console cmake application on Windows

Referring to the above topic of mine.
Have extended the code to this:

int main(int argc, char *argv[])
{
    int exitStatus = EXIT_SUCCESS;
    QScopedPointer<QCoreApplication> app( createApplication(argc, argv) );

    // if cast to QApp is successful => GUI mode was enabled
    if (qobject_cast<QApplication *>(app.data())) {

        //ShowWindow(GetConsoleWindow(), SW_HIDE);

        // start GUI version...
        MainWindow w;
        w.show();

        exitStatus = app->exec();

    } else {

        // start non-GUI version...

        // QStringList args = QCoreApplication::arguments();
        // parseArgs(args);

        BackendWorker worker;

        // Quit application when work is finished
        QObject::connect(&worker, &BackendWorker::finished, app.data(), &QCoreApplication::exit);  // [ 1 ]
        // QObject::connect(&worker, &BackendWorker::finished, app, &QCoreApplication::exit); // [ 2 ]

        QTimer::singleShot(0, &worker, &BackendWorker::start);

        exitStatus = app->exec(); // [ 3 ]
        // exitStatus = app.data()->exec(); // [ 4 ]

    }
    return exitStatus;
}

I'm confused about the connection part where [ 1 ] works and [ 2 ] does not.
Whereas both of [ 3 ] and [ 4 ] work and do the same.
And this is why it confuses me a little.

Uncommenting [ 2 ] throws:

error: no matching function for call to 'QObject::connect(BackendWorker *, void (BackendWorker::*)(int),
QScopedPointer<QCoreApplication>&, void (*)(int))'

69 | QObject::connect(&worker, &BackendWorker::finished, app, &QCoreApplication::exit);
~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Typing app->... even auto-completes/suggests functions like exec(), quit() or exit() (all part of QCoreApplication API)
I know these suggestions mean nothing :)

But if you are able to call QCoreApplication::exec on QScopedPointer<QCoreApplication> app as in [ 3 ], why you can't connect a signal like that (as in [ 2 ]) ?

Christian Ehrlicher

3 and 4 are the same because of this operator overload: https://doc.qt.io/qt-6/qscopedpointer.html#operator--gt

Pl45m4

@Christian-Ehrlicher said in QScopedPointer object, signal/slot usage, I am confused.:

3 and 4 are the same because of this operator overload: https://doc.qt.io/qt-6/qscopedpointer.html#operator--gt

lol, thank you, totally missed that.
Didn't even know that -> is an operator which you can overload :o
Same with . - operator (as I'm seeing it now)

Christian Ehrlicher

You can do nasty things with this. Overload operator + for integers to surprise your colleagues for example 😁