Qt connect lambda

Question

Hello I'm trying to connect to my QFrame component using lambda inside QMainWindow construct but I get error

Qwidget::mousePressEvent, cannot access protected member declared in class 'QWidget'

Here is my code

QtTestWindow::QtTestWindow(QWidget *parent):QMainWindow(parent)
{

connect( ui.TopFrame, &QFrame::mousePressEvent, [=]
{

});
ui.setupUi( this );

}

isn't it connect(&src, &signal(), &dst, &slot())? where is your destination? — user3528438
– user3528438, Commented Apr 4, 2015 at 23:12
@user3528438 that's the old syntax, the new one should be preferred — Alex Díaz
– Alex Díaz, Commented Apr 4, 2015 at 23:20
@AlejandroDíaz No it isn't. In both new and old there is a 4-argument and a 3-argument version. connect() 1 and connect() 3. The 3-argument version is "Equivalent to connect(sender, signal, this, method, type)." — Simon Warta
– Simon Warta, Commented Apr 5, 2015 at 7:56
@SimonWarta in that website there's an example showing the new syntax with a lambda function QObject::connect(socket, &QTcpSocket::connected, [=] () { socket->write("GET " + page + "\r\n"); }); , I think I'm missing your point — Alex Díaz
– Alex Díaz, Commented Apr 5, 2015 at 12:16

James Adkison · Accepted Answer · 2015-04-04 23:40:00Z

3

The QFrame class extents QWidget and the function is signature is

void QWidget::mousePressEvent(QMouseEvent * event) [virtual protected]

In other words this is not a signal and you cannot do what you're trying.

For completeness here's the documented signature of a signal

void QWidget::customContextMenuRequested(const QPoint & pos) [signal]

answered Apr 4, 2015 at 23:30

James Adkison

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