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

[Kai Krakow]

2018-06-28 12:39 GMT+02:00 Matteo Bruni <matteo.mystral at gmail.com>:
> 2018-06-28 9:05 GMT+02:00 Kai Krakow <kai at kaishome.de>:
>> Hello again!
>>
>> 2018-06-28 8:56 GMT+02:00 Kai Krakow <kai at kaishome.de>:
>>> 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.
>>
>> I just looked it up, the bang operator seems to be a logical (not
>> binary) operator in C99:
>> https://stackoverflow.com/questions/3661751/what-is-0-in-c
>>
>> According to this, I question how the code above is supposed to work
>> in the first place because "!modifiers[m].pressed" would always be 0
>> or 1, and it is compared to 0 or 0x80.
>
> No, it's compared to 0 or 1 (there is a ! on the left hand side of the == too)

Oh *facepalm*, you're right. I didn't notice the opening and closing
parentheses in the correct position.

>> This is exploiting implicit
>> behavior which is almost always a bad thing to do if you need to
>> understand code later.
>
> It's pretty clear and explicit I'd say.

Maybe but still not very visible...

But this should answer the OPs question why it is done in this way.


Thanks for clarification,
Kai