XBMC might be different as it handles events itself but generally speaking
you should repeat the release event as a keystroke is made out of press and
release.
On 28/10/2013 3:31 pm, "Rainer Hochecker" <notifications@github.com> wrote:

> In xbmc/windowing/WinEventsX11.cpp:
>
> > +        break;
> > +      } //KeyPress
> > +
> > +      case KeyRelease:
> > +      {
> > +        // if we have a queued press directly after, this is a repeat
> > +        if( XEventsQueued( WinEvents->m_display, QueuedAfterReading ) )
> > +        {
> > +          XEvent next_event;
> > +          XPeekEvent( WinEvents->m_display, &next_event );
> > +          if(next_event.type == KeyPress
> > +            && next_event.xkey.window == xevent.xkey.window
> > +            && next_event.xkey.keycode == xevent.xkey.keycode
> > +            && (next_event.xkey.time - xevent.xkey.time < 2) )
> > +            continue;
> > +        }
>
> Doesn't this respect the system setting? In case of a key repeat the
> application should not be notified about the release event.
>
> —
> Reply to this email directly or view it on GitHub<https://github.com/xbmc/xbmc/pull/3505/files#r7241343>
> .
>