Selecting screen on the application starts

Hi,

Has Qt any functions to enumerate the displays on a system and to force an application to start on the first or second display?

Thanks,

QApplication::desktop ()->numScreens () gets number of screens. I suppose that you should keep digging QDesktopWidget reference to do what you want

Hi,

Thanks, I was searching for "display", "screen" and had not think on "desktop".

Hi,

The problem is, how can I force my application to start to the second display?

I'm creating an application that have no frame, so it can't be moved by mouse to the other display.

Thanks,

Hi,

Code:

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myMainWindow mainWindow(app.desktop()->screen(iScreen));

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I can take the desktop as a widget an set it as the parent of my QMainWindow.

Thanks,

Hi,

I have a problem,
The problem is that "isVirtualDesktop" is returning me true, and the number of screens return me 2.

Code:

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bool b = app.desktop()->isVirtualDesktop();  //true
QWidget* pDesktop0 = app.desktop()->screen(0);
QWidget* pDesktop1 = app.desktop()->screen(1);

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pDesktop0 and pDesktop1 point to the same memory data.

Hi,

Finally I solved this problem on Virtual Desktops:

Code:

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QRect qAvGeometry = app.desktop()->availableGeometry(0);
//Move the MainWindow to left by horizontal pixels(horizontal resolution)
mainWindow.move((iScreen-1)*(qAvGeometry.width()+1),0);

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