Re: [Qemu-devel] Patch: cpu-all.h for better cycle counter

["andrzej zaborowski" <balrog@zabor.org>]

On 26/03/07, Marc Lörner <marc.loerner@hob.de> wrote:
> On Monday 26 March 2007 00:08, andrzej zaborowski wrote:
> > On 26/03/07, andrzej zaborowski <balrog@zabor.org> wrote:
> <snip>
> > > > +# warning non-optimized CPU
> > > > +#include <sys/time.h>
> > > > +#include <time.h>
> > > > +
> > > >  static inline int64_t cpu_get_real_ticks (void)
> > > >  {
> > > > -    static int64_t ticks = 0;
> > > > -    return ticks++;
> > > > +       struct timeval tv;
> > > > +       static int64_t i = 0;
> > > > +       int64_t j;
> > > > +
> > > > +       gettimeofday(&tv, NULL);
> > > > +       do {
> > > > +          j = (tv.tv_sec * (uint64_t) 1000000) + tv.tv_usec;
> > > > +       } while (i == j);
> > > > +       i = j;
> > > > +       return j;
> > >
> > > Isn't this an infinite loop? gettimeofday() was left out of the loop.
> > >
> > > How about "return j + (ticks++)" instead of the loop? If I understand
> > > correctly it may slow things down to below 1Hz.
> >
> > (I wanted to say MHz)
>
> I dont think so, in the loop "j" is set and the while-condition is "j==i",
> so unless "(tv.tv_sec * (uint64_t) 1000000) + tv.tv_usec" always computes the
> same value the do-while block gets only executed once.

Well, it does always compute the same value - inside the loop. Doesn't
it? And across one microsecond also the same value in all calls to
cpu_get_real_ticks.

<snip>

Regards