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Windows: Unstick shift keys in a timely manner when the OS loses a KEYUP event.

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Fixes Bugzilla #1959.
This commit is contained in:
Ryan C. Gordon 2013-08-01 01:50:02 -04:00
parent 9b622dc41e
commit 72d8cda7bb
1 changed files with 13 additions and 0 deletions
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13
src/video/windows/SDL_windowsevents.c
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@ -766,11 +766,24 @@ WIN_WindowProc(HWND hwnd, UINT msg, WPARAM wParam, LPARAM lParam)
void
WIN_PumpEvents(_THIS)
{
const Uint8 *keystate;
MSG msg;
while (PeekMessage(&msg, NULL, 0, 0, PM_REMOVE)) {
TranslateMessage(&msg);
DispatchMessage(&msg);
}
/* Windows loses a shift KEYUP event when you have both pressed at once and let go of one.
You won't get a KEYUP until both are released, and that keyup will only be for the second
key you released. Take heroic measures and check the keystate as of the last handled event,
and if we think a key is pressed when Windows doesn't, unstick it in SDL's state. */
keystate = SDL_GetKeyboardState(NULL);
if ((keystate[SDL_SCANCODE_LSHIFT] == SDL_PRESSED) && !(GetKeyState(VK_LSHIFT) & 0x8000)) {
SDL_SendKeyboardKey(SDL_RELEASED, SDL_SCANCODE_LSHIFT);
}
if ((keystate[SDL_SCANCODE_RSHIFT] == SDL_PRESSED) && !(GetKeyState(VK_RSHIFT) & 0x8000)) {
SDL_SendKeyboardKey(SDL_RELEASED, SDL_SCANCODE_RSHIFT);
}
}
static int app_registered = 0;