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Fix creating disabled windows in wxGTK too

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Don't forbid calling Enable() before creating the window and just do
nothing in this case and do disable the window when it's actually
created if it's supposed to be disabled.

Note that this doesn't work for classes overriding Enable() directly,
such as wxButton or wxCheckBox, currently.
This commit is contained in:
Vadim Zeitlin
2018-12-09 01:15:08 +01:00
parent dfec7aa0c0
commit 5ba1ba1162
1 changed files with 11 additions and 1 deletions
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12
src/gtk/window.cpp
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@@ -2913,6 +2913,11 @@ void wxWindowGTK::PostCreation()
SetLayoutDirection(wxLayout_Default); SetLayoutDirection(wxLayout_Default);
// if the window had been disabled before being created, it should be
// created in the initially disabled state
if ( !m_isEnabled )
DoEnable(false);
// unless the window was created initially hidden (i.e. Hide() had been // unless the window was created initially hidden (i.e. Hide() had been
// called before Create()), we should show it at GTK+ level as well // called before Create()), we should show it at GTK+ level as well
if (m_isShown) if (m_isShown)
@@ -4196,7 +4201,12 @@ bool wxWindowGTK::IsShown() const
void wxWindowGTK::DoEnable( bool enable ) void wxWindowGTK::DoEnable( bool enable )
{ {
wxCHECK_RET( (m_widget != NULL), wxT("invalid window") ); if ( !m_widget )
{
// The window can be disabled before being created, so just don't do
// anything in this case and, in particular, don't assert.
return;
}
gtk_widget_set_sensitive( m_widget, enable ); gtk_widget_set_sensitive( m_widget, enable );
if (m_wxwindow && (m_wxwindow != m_widget)) if (m_wxwindow && (m_wxwindow != m_widget))