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input: send kbd+mouse events only to running guests. 

Trying to interact with a stopped guest will queue up the events,
then send them all at once when the guest continues running, with
a high chance to have them cause unwanted actions.

Avoid that by only injecting the input events only when the guest
is in running state.

Signed-off-by: Gerd Hoffmann <kraxel@redhat.com>
Signed-off-by: Anthony Liguori <aliguori@us.ibm.com>
This commit is contained in:
Gerd Hoffmann 2012-02-15 09:15:37 +01:00 committed by Anthony Liguori
parent aa24822bdc
commit 99c7f87826
1 changed files with 6 additions and 0 deletions
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6
input.c
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@ -130,6 +130,9 @@ void qemu_remove_led_event_handler(QEMUPutLEDEntry *entry)
void kbd_put_keycode(int keycode)
{
if (!runstate_is_running()) {
return;
}
if (qemu_put_kbd_event) {
qemu_put_kbd_event(qemu_put_kbd_event_opaque, keycode);
}
@ -151,6 +154,9 @@ void kbd_mouse_event(int dx, int dy, int dz, int buttons_state)
void *mouse_event_opaque;
int width, height;
if (!runstate_is_running()) {
return;
}
if (QTAILQ_EMPTY(&mouse_handlers)) {
return;
}