[SOLVED] How to pass a signal forwarded parameter into a lambda, defined in the connect statement?

utcenter

I am having difficulties figuring out how to get access to a parameter, passed by a signal in the lambda the signal is connected to. Any ideas?

Chris Kawa

You mean something like this?

connect(someButton, &QPushButton::toggled, [](bool checked) { if(checked) .... ; });

utcenter

Ah yes, it worked, I wasn't getting any autocomplete so I figured I must have messed something up. Gotta try upgrading to the new Creator.

QObject::connect(&b, static_cast<void (Widget::*)(QMouseEvent *)>(&Widget::clicked), [=](QMouseEvent *ev){
            if (ev->button() == Qt::RightButton) qDebug() << "right";
        });

Chris Kawa

What's the static_cast for? Wouldn't this suffice?

QObject::connect(&b, &Widget::clicked, [](QMouseEvent *ev){
            if (ev->button() == Qt::RightButton) qDebug() << "right";
        });

utcenter

Last time I used an overloaded signal or slot the compiler complained it cannot resolve the methods. So I had to use this awkward syntax to get the address of the right overload, otherwise it wouldn't compile.

I didn't expect this to work at all. But if get it correctly, the lambda parameter helped resolve the right address of clicked, or it just blindly connects to the derived signal instead the one in the base class, but then how does the lambda parameter know where to come from?