From df0f099b7ae0ffbea46411276a239d152f36a7e2 Mon Sep 17 00:00:00 2001 From: Gareth Stockwell Date: Tue, 20 Oct 2009 17:50:11 +0100 Subject: When creating a Symbian WId for a visible widget, make the control visible after activating the window. This change was required in order to be able to run the test case for the below task; however, it is more generally required. Without it, the contents of the descendents of this widget will not be visible, until they are explicitly hidden and then re-shown. Task-number: QTBUG-4787 Reviewed-by: axis --- src/gui/kernel/qwidget_s60.cpp | 4 +++- 1 file changed, 3 insertions(+), 1 deletion(-) diff --git a/src/gui/kernel/qwidget_s60.cpp b/src/gui/kernel/qwidget_s60.cpp index cb615fe..3e332a3 100644 --- a/src/gui/kernel/qwidget_s60.cpp +++ b/src/gui/kernel/qwidget_s60.cpp @@ -434,8 +434,10 @@ void QWidgetPrivate::create_sys(WId window, bool /* initializeWindow */, bool de drawableWindow->PointerFilter(EPointerFilterEnterExit | EPointerFilterMove | EPointerFilterDrag, 0); - if (q->isVisible() && q->testAttribute(Qt::WA_Mapped)) + if (q->isVisible() && q->testAttribute(Qt::WA_Mapped)) { activateSymbianWindow(control.data()); + control->MakeVisible(true); + } // We wait until the control is fully constructed before calling setWinId, because // this generates a WinIdChanged event. -- cgit v0.12