提交 562f9375 编写于 作者: P Paolo Bonzini 提交者: Gerd Hoffmann

hid: clarify hid_keyboard_process_keycode

Coverity thinks the fallthroughs are smelly.  They are correct, but
everything else in this function is like "wut?".

Refer explicitly to bits 8 and 9 of hs->kbd.modifiers instead of
shifting right first and using (1 << 7).  Document what the scancode
is when hid_code is 0xe0.  And add plenty of comments.
Signed-off-by: NPaolo Bonzini <pbonzini@redhat.com>
Signed-off-by: NGerd Hoffmann <kraxel@redhat.com>
上级 e2f6bac3
......@@ -239,7 +239,7 @@ static void hid_keyboard_event(DeviceState *dev, QemuConsole *src,
static void hid_keyboard_process_keycode(HIDState *hs)
{
uint8_t hid_code, key;
uint8_t hid_code, index, key;
int i, keycode, slot;
if (hs->n == 0) {
......@@ -249,7 +249,8 @@ static void hid_keyboard_process_keycode(HIDState *hs)
keycode = hs->kbd.keycodes[slot];
key = keycode & 0x7f;
hid_code = hid_usage_keys[key | ((hs->kbd.modifiers >> 1) & (1 << 7))];
index = key | ((hs->kbd.modifiers & (1 << 8)) >> 1);
hid_code = hid_usage_keys[index];
hs->kbd.modifiers &= ~(1 << 8);
switch (hid_code) {
......@@ -257,18 +258,41 @@ static void hid_keyboard_process_keycode(HIDState *hs)
return;
case 0xe0:
assert(key == 0x1d);
if (hs->kbd.modifiers & (1 << 9)) {
hs->kbd.modifiers ^= 3 << 8;
/* The hid_codes for the 0xe1/0x1d scancode sequence are 0xe9/0xe0.
* Here we're processing the second hid_code. By dropping bit 9
* and setting bit 8, the scancode after 0x1d will access the
* second half of the table.
*/
hs->kbd.modifiers ^= (1 << 8) | (1 << 9);
return;
}
/* fall through to process Ctrl_L */
case 0xe1 ... 0xe7:
/* Ctrl_L/Ctrl_R, Shift_L/Shift_R, Alt_L/Alt_R, Win_L/Win_R.
* Handle releases here, or fall through to process presses.
*/
if (keycode & (1 << 7)) {
hs->kbd.modifiers &= ~(1 << (hid_code & 0x0f));
return;
}
case 0xe8 ... 0xef:
/* fall through */
case 0xe8 ... 0xe9:
/* USB modifiers are just 1 byte long. Bits 8 and 9 of
* hs->kbd.modifiers implement a state machine that detects the
* 0xe0 and 0xe1/0x1d sequences. These bits do not follow the
* usual rules where bit 7 marks released keys; they are cleared
* elsewhere in the function as the state machine dictates.
*/
hs->kbd.modifiers |= 1 << (hid_code & 0x0f);
return;
case 0xea ... 0xef:
abort();
default:
break;
}
if (keycode & (1 << 7)) {
......
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