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>

1 files changed, 6 insertions, 0 deletions

diff --git a/input.c b/input.c
index b618ea4c1c..6b5c2c3371 100644
--- a/input.c
+++ b/input.c
@@ -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;
     }