Make sure that QAccessibleWindowContainer::childCount is valid

When embedding foreign windows, we won't be able to return a valid child accessible interface, so do not report it at all. Supporting foreign windows properly is platform specific and something to consider, but at least we shouldn't crash. Task-number: QTBUG-63451 Change-Id: I19350cf97dc8d0c3f3052411eba0eee5f750dbab Reviewed-by: Jan Arve Sæther <[email protected]>
tmpsantos · Oct 25, 2017 · 7a26582 · 7a26582
1 parent 5eb508a
commit 7a26582
Showing 1 changed file with 1 addition and 1 deletion.
diff --git a/src/widgets/accessible/simplewidgets.cpp b/src/widgets/accessible/simplewidgets.cpp
@@ -953,7 +953,7 @@ QAccessibleWindowContainer::QAccessibleWindowContainer(QWidget *w)
 
 int QAccessibleWindowContainer::childCount() const
 {
-    if (container()->containedWindow())
+    if (container()->containedWindow() && QAccessible::queryAccessibleInterface(container()->containedWindow()))
         return 1;
     return 0;
 }