On Tue, 16 Apr 2013 09:08:12 +0200
Markus Armbruster <arm...@redhat.com> wrote:
> Amos Kong <ak...@redhat.com> writes:
>
> > (qemu) sendkey ctrl_r-scroll_lock-scroll_lock
> >
> > Executing this command could not let Windows guest panic, it caused by
>
> "could not"?
You can use Windows' editor to setup Windows to panic on a key sequence, and
that feature (or Windows itself) is sensible to the key release ordering.
This caught a problem with the sendkey command conversion to the qapi: we
changed the release key ordering from last to first to first to last.
> > the wrong order of releasing keys. This problem was introduced by
> > commit e4c8f004c55d9da3eae3e14df740238bf805b5d6.
> >
> > The right release order should be starting from last item.
> >
> > Signed-off-by: Amos Kong <ak...@redhat.com>
> > ---
> > ui/input.c | 8 +++-----
> > 1 files changed, 3 insertions(+), 5 deletions(-)
> >
> > diff --git a/ui/input.c b/ui/input.c
> > index 9abef0c..ecfeb43 100644
> > --- a/ui/input.c
> > +++ b/ui/input.c
> > @@ -234,13 +234,11 @@ static void free_keycodes(void)
> >
> > static void release_keys(void *opaque)
> > {
> > - int i;
> > -
> > - for (i = 0; i < keycodes_size; i++) {
> > - if (keycodes[i] & 0x80) {
> > + while (keycodes_size > 0) {
> > + if (keycodes[--keycodes_size] & 0x80) {
> > kbd_put_keycode(0xe0);
> > }
> > - kbd_put_keycode(keycodes[i]| 0x80);
> > + kbd_put_keycode(keycodes[keycodes_size] | 0x80);
> > }
> >
> > free_keycodes();
>
