Interrupt performance

[Slawa Olhovchenkov]

On Sat, Jan 29, 2011 at 09:55:06PM -0800, Robby Sun wrote:

> I'd like to suggest that you use the same bit-width for 'Dummy' as that for
> 'count', and initialize it to 0, so as to ensure that it won't overflow.

I don't use value of Dummy, overflow don't meaning.

> -Robby
> 
> On Fri, Jan 28, 2011 at 1:52 PM, Slawa Olhovchenkov <slw at zxy.spb.ru> wrote:
> 
> > On Sat, Jan 29, 2011 at 07:52:11AM +1100, Bruce Evans wrote:
> >
> > > >> there are of course several possible answers, including:
> > > >>
> > > >> 1/ Sometimes BSD and Linux report things differently. Linux may or may
> > not
> > > >> account for the lowest level interrupt tie the same as BSD
> > > >
> > > > But I see only 20% idle on FreeBSD and 80% idle on Linux.
> > >
> > > The time must be counted somewhere, so when it is not properly accounted
> > > to packet handling, and nothing much else is running, it is accounted to
> > > idle.
> > >
> > > To see how much CPU is actually available, run something else and see how
> > > fast it runs.  A simple counting loops works well on UP systems.
> >
> > ===
> > #include <stdio.h>
> > #include <sys/time.h>
> >
> > int Dummy;
> >
> > int
> > main(int argc, char *argv[])
> > {
> >  long int count,i,dt;
> >  struct timeval st,et;
> >
> >  count = atol(argv[1]);
> >
> >  gettimeofday(&st, NULL);
> >  for(i=count;i;i--) Dummy++;
> >  gettimeofday(&et, NULL);
> >  dt = (et.tv_sec-st.tv_sec)*1000000 + et.tv_usec-st.tv_usec;
> >  printf("Elapsed %d us\n",dt);
> > }
> > ===
> >
> > This is ok?
> >
> > ./loop 2000000000
> >
> > FreeBSD
> > 1 process: Elapsed 7554193 us
> > 2 process: Elapsed 14493692 us
> > netperf + 1 process: Elapsed 21403644 us
> >
> > Linux
> > 1 process: Elapsed 7524843 us
> > 2 process: Elapsed 14995866 us
> > netperf + 1 process: Elapsed 14107670 us
> >
> >
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