Always open a submenu on the same screen where the parent menu is

We've seen submenus opening on different screens in different scenarios,
but probably there's never a sensible reason to do that.
QWidgetPrivate::setScreenForPoint() can often give incorrect results
in multi-screen desktop configurations; this is a way of avoiding that.

Task-number: QTBUG-76162
Change-Id: I895696c6858ad953258c84b21117beea4c4fc01d
Reviewed-by: Andy Shaw <andy.shaw@qt.io>

1 files changed, 11 insertions, 2 deletions

diff --git a/src/widgets/widgets/qmenu.cpp b/src/widgets/widgets/qmenu.cpp
index 865e3b2fb6..38fdbf2184 100644
--- a/src/widgets/widgets/qmenu.cpp
+++ b/src/widgets/widgets/qmenu.cpp
@@ -40,6 +40,7 @@
 #include "qmenu.h"
 
 #include <QtWidgets/private/qtwidgetsglobal_p.h>
+#include <QtWidgets/private/qwidgetwindow_p.h>
 
 #include "qdebug.h"
 #include "qstyle.h"
@@ -2343,15 +2344,23 @@ void QMenuPrivate::popup(const QPoint &p, QAction *atAction, PositionFunction po
     // Use d->popupScreen to remember, because initialScreenIndex will be reset after the first showing.
     // However if eventLoop exists, then exec() already did this by calling createWinId(); so leave it alone. (QTBUG-76162)
     if (!eventLoop) {
+        bool screenSet = false;
         const int screenIndex = topData()->initialScreenIndex;
         if (screenIndex >= 0)
             popupScreen = screenIndex;
         if (auto s = QGuiApplication::screens().value(popupScreen)) {
             if (setScreen(s))
                 itemsDirty = true;
-        } else if (setScreenForPoint(p)) {
-            itemsDirty = true;
+            screenSet = true;
+        } else if (QMenu *parentMenu = qobject_cast<QMenu *>(parent)) {
+            // a submenu is always opened from an open parent menu,
+            // so show it on the same screen where the parent is. (QTBUG-76162)
+            if (setScreen(QMenuPrivate::get(parentMenu)->windowHandle()->screen()))
+                itemsDirty = true;
+            screenSet = true;
         }
+        if (!screenSet && setScreenForPoint(p))
+            itemsDirty = true;
     }
 
     const bool contextMenu = isContextMenu();