From: mauch AT uni-duisburg DOT de (Michael Mauch)
Newsgroups: comp.os.msdos.djgpp
Subject: Re: testing uclock()
Date: Mon, 05 May 1997 12:31:19 +0200
Organization: Home, sweet home (via Gesamthochschule Duisburg)
Lines: 131
Distribution: world
Message-ID: <336ea503.189123@news.uni-duisburg.de>
References: <Pine DOT SUN DOT 3 DOT 91 DOT 970504133340 DOT 9215E-100000 AT is>
NNTP-Posting-Host: ppp91.uni-duisburg.de
To: djgpp AT delorie DOT com
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On Sun, 4 May 1997 10:34:10 GMT, Eli Zaretskii <eliz AT is DOT elta DOT co DOT il>
wrote:

> > On the other hand: isn't it possible that this default mode is just
> > the same as the one set by the first call to uclock()? The
> > difference between two consecutive calls to uclock() remains the
> > same, even after that single negative difference.
> 
> No, you can't work with the default mode at all.  Under that mode, the
> timer counter goes up and down twice during the 54 msec period, so you
> cannot reliably compute the interval.

Oh, that's bad.
 
> > I removed the first three outportb() calls from the uclock source
> > code, and now I don't get any negative differences anymore.
> 
> I think that is specific to the time that your test program waited
> between two calls to `uclock'.  Try different intervals and you will
> see the problem again.

Yes, you are right, unfortunately.

In the meantime I translated my 16 bit source code to work with DJGPP. 
It uses the Virtual Timer Device of Win3.1/Win95 if this is available,
otherwise it calls the original uclock(). Although it seems to work, I'm
not sure how accurate it is - as much as we can expect timing accuracy
in Windows. I looked at the output of a test program (with different
intervals, this time) using gnuplot, and it shows up that it is a lot
slower than uclock(), and there are a lot of peaks (some of them up to
60000 tics, i.e. 50ms), but there a no negative values anymore.

Another question is, if there is a wrap-around anywhere (there's none at
midnight). Ralf Brown's Interrupt List says that the VTD API gives the
time since Windows was started, so there should be a long time before
the 63 bit signed long long (returned in edx:eax by the VTD API call)
would need a wrap around. But you never know.

And I don't know why the original uclock() source uses __bss_count (what
is __bss_count?):

  if (uclock_bss != __bss_count)
  {
    /* timer chip initialization here */
    uclock_bss = __bss_count;
  }

Is it simply a flag to do the initialization only once or what's the
magic with __bss_count?

Regards...
		Michael


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