From: "Marvin Gülker" <sutniuq@...>
Date: 2009-12-30T22:02:48+09:00
Subject: Re: au3 0.1.1 released

Heesob Park wrote:
> As you know, the INPUT structure defined like this:
> 
> typedef struct tagINPUT {
>   DWORD type;
>   union {MOUSEINPUT mi;
>             KEYBDINPUT ki;
>             HARDWAREINPUT hi;
>            };
>   }INPUT, *PINPUT;
> 
> typedef struct tagMOUSEINPUT {
>     LONG dx;
>     LONG dy;
>     DWORD mouseData;
>     DWORD dwFlags;
>     DWORD time;
>     ULONG_PTR dwExtraInfo;
> } MOUSEINPUT, *PMOUSEINPUT;
> 
> typedef struct tagKEYBDINPUT {
>     WORD wVk;
>     WORD wScan;
>     DWORD dwFlags;
>     DWORD time;
>     ULONG_PTR dwExtraInfo;
> } KEYBDINPUT, *PKEYBDINPUT;
> 
> typedef struct tagHARDWAREINPUT {
>     DWORD uMsg;
>     WORD wParamL;
>     WORD wParamH;
> } HARDWAREINPUT, *PHARDWAREINPUT;
> 
> The sizeof(MOUSEINPUT) is 24, sizeof(KEYBDINPUT) is 16, and
> sizeof(HARDWAREINPUT) is 8.
> Therefore, sizeof(INPUT) = sizeof(DWORD) + maxsizeof(union) = 4 + 24 = 
> 28.
> 
> Regards,
> 
> Park Heesob

Thank you again, now I've got the point. One last question: The size of 
the MOUSEINPUT struct is 8 + 8 + 4 + 4 + 4 = 28 I see, so am I right in 
thinking that the pointer isn't important for this computation? 
Otherwise, since it's an unsigned long, it should have a size of 8 also, 
what makes the struct a size of 36 bytes? Or is this ignored because of 
passing in nil later?

You see, I'm not really good in working with C structs. :)

Marvin
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