Find the active QScreen of any QWidget

Richard

Hi, Is there an easy way to find a widget's active QScreen it resides on, obviously in a multi display environment. I am using Qt5.8 on windows 10.

I have extended the desktop with with an additional monitor which has lower resolution then my laptop.
This gives me some scaling issues with QPicture I am trying to work around.
I tried the function windowsHandle() which has a function screen but this function returns zero, propably because my widget is not a top level widget but instead part of an MDI child.

mrjj

Hi
Could you not you use
http://doc.qt.io/qt-5/qdesktopwidget.html#screenNumber
"Returns the index of the screen that contains the largest part of widget"

Richard

Hi, Thanks for your suggestion. I tried this but this always returns 0. This is correct though, I extended my desktop so there is always one virtual screen.
But then again QDesktopWidget::screenCount() return 2. So this is a bit odd.

Richard

I found this in the docu:

"virtualDesktop : const bool
This property holds if the system manages the available screens in a virtual desktop.
For virtual desktops, screen() will always return the same widget. The size of the virtual desktop is the size of this desktop widget."

So... yep virtualDeskTop() returns true.

mrjj

Hi
Ok, i think you need to loop over the screens and their geometry and
check that against the apps x,y to see on which screen it is placed.
if its x are bigger than screen 0 width, then its on screen 1.

Notice, that apps. x might be negative. At least thats the case when placing
remote desktop windows and second monitor is on the left. . I didn't test
what Qt reports in this case.

Technically its one big screen so hence screenNumber reports correctly.
Sorry, i missed the "extended" word in your description. :)

Richard

This works ok. But not every widget has a windowHandle (maybe olny top level ones?) So then i grab the parent to check it.

QScreen* getActiveScreen(QWidget* pWidget) const
		{
			QScreen* pActive = nullptr;

			while (pWidget)
			{
				auto w = pWidget->windowHandle();
				if (w != nullptr)
				{
					pActive = w->screen();
					break;
				}
				else
					pWidget = pWidget->parentWidget();
			}

			return pActive;
		}

mrjj

Hi
ok so for a top level widget/window, ->screen()
actually knows which screen even if its a virtual screen/extended.
That is good to know :)
Its explained here whom have >windowHandle();
http://doc.qt.io/qt-5/qwidget.html#windowHandle