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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: Paolo Bonzini <pbonzini@redhat.com> Signed-off-by: Gerd Hoffmann <kraxel@redhat.com>
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1 changed files with 28 additions and 4 deletions
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@ -239,7 +239,7 @@ static void hid_keyboard_event(DeviceState *dev, QemuConsole *src,
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static void hid_keyboard_process_keycode(HIDState *hs)
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{
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uint8_t hid_code, key;
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uint8_t hid_code, index, key;
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int i, keycode, slot;
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if (hs->n == 0) {
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@ -249,7 +249,8 @@ static void hid_keyboard_process_keycode(HIDState *hs)
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keycode = hs->kbd.keycodes[slot];
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key = keycode & 0x7f;
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hid_code = hid_usage_keys[key | ((hs->kbd.modifiers >> 1) & (1 << 7))];
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index = key | ((hs->kbd.modifiers & (1 << 8)) >> 1);
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hid_code = hid_usage_keys[index];
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hs->kbd.modifiers &= ~(1 << 8);
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switch (hid_code) {
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@ -257,18 +258,41 @@ static void hid_keyboard_process_keycode(HIDState *hs)
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return;
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case 0xe0:
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assert(key == 0x1d);
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if (hs->kbd.modifiers & (1 << 9)) {
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hs->kbd.modifiers ^= 3 << 8;
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/* The hid_codes for the 0xe1/0x1d scancode sequence are 0xe9/0xe0.
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* Here we're processing the second hid_code. By dropping bit 9
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* and setting bit 8, the scancode after 0x1d will access the
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* second half of the table.
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*/
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hs->kbd.modifiers ^= (1 << 8) | (1 << 9);
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return;
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}
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/* fall through to process Ctrl_L */
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case 0xe1 ... 0xe7:
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/* Ctrl_L/Ctrl_R, Shift_L/Shift_R, Alt_L/Alt_R, Win_L/Win_R.
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* Handle releases here, or fall through to process presses.
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*/
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if (keycode & (1 << 7)) {
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hs->kbd.modifiers &= ~(1 << (hid_code & 0x0f));
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return;
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}
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case 0xe8 ... 0xef:
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/* fall through */
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case 0xe8 ... 0xe9:
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/* USB modifiers are just 1 byte long. Bits 8 and 9 of
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* hs->kbd.modifiers implement a state machine that detects the
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* 0xe0 and 0xe1/0x1d sequences. These bits do not follow the
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* usual rules where bit 7 marks released keys; they are cleared
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* elsewhere in the function as the state machine dictates.
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*/
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hs->kbd.modifiers |= 1 << (hid_code & 0x0f);
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return;
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case 0xea ... 0xef:
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abort();
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default:
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break;
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}
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if (keycode & (1 << 7)) {
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