Bug #45385 related to keyboard and probably wineserver - where to start the search

[Kai Krakow]

Hello!

2018-06-27 21:14 GMT+02:00 Alex Henrie <alexhenrie24 at gmail.com>:
> On Wed, Jun 27, 2018 at 11:24 AM John Found <johnfound at asm32.info> wrote:
>>
>> While reading the code in wineX11.drv/keyboard.c I noticed the following fragment from the function X11DRV_KeymapNotify:
>>
>>     for (vkey = VK_LSHIFT; vkey <= VK_RMENU; vkey++)
>>     {
>>         int m = vkey - VK_LSHIFT;
>> >>>     if (modifiers[m].vkey && !(keystate[vkey] & 0x80) != !modifiers[m].pressed)
>>         {
>>             TRACE( "Adjusting state for vkey %#.2x. State before %#.2x\n",
>>                    modifiers[m].vkey, keystate[vkey]);
>>
>>             update_key_state( keystate, vkey, modifiers[m].pressed );
>>             changed = TRUE;
>>         }
>>     }
>>
>> Can someone explain what the condition pointed by >>> is doing? (Unfortunately, my C skills are pretty low...)
>>
>> Why not simply "(modifiers[m].vkey && (keystate[vkey] & 0x80) != modifiers[m].pressed)"?
>
> I don't immediately know what this if statement is for, but I can tell
> you that the ! operator converts a value of zero to 1 and a nonzero
> value to 0. So, if keystate[vkey] & 0x80 is 0 then
> modifiers[m].pressed must be nonzero and vice versa.

I always was under the impression that the bang operator does a binary
inversion of the value. Thus, it would convert 0x7F to 0x80. This also
means it would convert 0x00 (signed) to 0xFF (signed) which turns 0
into -1, not 1.

But however it works, this code is not working in any obvious way and
should maybe be rewritten.

Just my two cents.

Regards,
Kai