Vojtech Pavlik wrote:
>
> On Fri, Dec 20, 2002 at 09:24:58AM -0800, george anzinger wrote:
> > Vojtech Pavlik wrote:
> > >
> > > On Fri, Dec 20, 2002 at 02:41:50AM -0800, george anzinger wrote:
> > > > Bjorn Helgaas wrote:
> > > > >
> > > > > * fix a problem with HZ->millisecond transformation on
> > > > > non-x86 archs (from 2.5 change by vojtech@suse.cz)
> > > > >
> > > > > Applies to 2.4.20.
> > > > >
> > > > > diff -Nru a/drivers/input/joydev.c b/drivers/input/joydev.c
> > > > > --- a/drivers/input/joydev.c Mon Dec 16 12:16:32 2002
> > > > > +++ b/drivers/input/joydev.c Mon Dec 16 12:16:32 2002
> > > > > @@ -50,6 +50,8 @@
> > > > > #define JOYDEV_MINORS 32
> > > > > #define JOYDEV_BUFFER_SIZE 64
> > > > >
> > > > > +#define MSECS(t) (1000 * ((t) / HZ) + 1000 * ((t) % HZ) / HZ)
> > > > Uh...
> > > > ^^^^^^^^^^^^^^^^
> > > > by definition this is zero, is it not?
> > >
> > > No, both parts of the equaition can be nonzero.
> >
> > I don't think so. s%HZ has to be less than HZ. Then
> > dividing that by HZ should result in zero. Where is my
> > thinking flawed?
>
> You first multiply it by 1000.
But then it should read: (1000 * (t)) % HZ) / HZ
But this is still zero. There is no way that ((X % HZ) /
HZ) is other than zero, me thinks.
>
> > > Though it might be easier to say (1000 * t) / HZ, now that I think about
> > > it.
> >
> > That overflows... As does the other if HZ is less than 1000....
>
> You're right, t can be all 32 bits.
What, exactly, is this code trying to do? It looks like the
number is being passed to user space...
>
> --
> Vojtech Pavlik
> SuSE Labs
> -
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-- George Anzinger george@mvista.com High-res-timers: http://sourceforge.net/projects/high-res-timers/ Preemption patch: http://www.kernel.org/pub/linux/kernel/people/rml - To unsubscribe from this list: send the line "unsubscribe linux-kernel" in the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html Please read the FAQ at http://www.tux.org/lkml/
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