[RFC 5/11] Linux FF: Linux effect parameters: time parameters

[Andrew Eikum]

On Tue, Mar 04, 2014 at 08:20:09PM +0100, Elias Vanderstuyft wrote:
>     So change in dinput/effect_linuxinput.c:488 :
>         This->effect.u.periodic.period = tsp->dwPeriod / 1000;
>     to :
>         if (0 < tsp->dwPeriod && tsp->dwPeriod <= 1000)
>             This->effect.u.periodic.period = 1;
>         else
>             This->effect.u.periodic.period = tsp->dwPeriod / 1000;
> 

Seems good. While we're here, any reason not to use round(3) instead
of always rounding down in the >1000 case?

> b)  Anticipate if dwPeriod < 0, otherwise periodic.period will
> incorrectly wrap around because it's unsigned.
>     We can choose to let periodic.period be proportional to:
>     - 0    (driver will generate EINVAL)
>     - abs(dwPeriod)    (no error, and is mathematically more correct)

DWORDs are usually treated as unsigned. Is this how Windows behaves
for massive dwPeriod? I would expect E_INVALIDARG, since it's out of
the range [0,36000).

>     So, change in dinput/effect_linuxinput.c:191 :
>         if (dwFlags & DIEP_DURATION) {
>             peff->dwDuration = (DWORD)This->effect.replay.length * 1000;
>         }
>     to :
>         if (dwFlags & DIEP_DURATION) {
>             peff->dwDuration = (This->effect.replay.length ?
> (DWORD)This->effect.replay.length * 1000 : -1);
>         }
> 
>     And change in dinput/effect_linuxinput.c:424 :
>         if (dwFlags & DIEP_DURATION)
>             This->effect.replay.length = peff->dwDuration / 1000;
>     to :
>         if (dwFlags & DIEP_DURATION) {
>             if (peff->dwDuration == -1)
>                 This->effect.replay.length = 0; // Infinity
>             else if (peff->dwDuration > 1000)
>                 This->effect.replay.length = peff->dwDuration / 1000;
>             else
>                 This->effect.replay.length = 1; // Smallest non-zero value
>         }
> 

This seems okay, though again, maybe use round(). You should also use
INFINITE instead of -1.

Andrew