linux-kernel - Re: [PATCH][resubmit] x86: enable preemption in delay

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Date:	Sun, 1 Jun 2008 20:17:57 +0300
From:	Marin Mitov <mitov@...p.bas.bg>
To:	Andi Kleen <andi@...stfloor.org>
Cc:	LKML <linux-kernel@...r.kernel.org>,
	Steven Rostedt <rostedt@...dmis.org>,
	Thomas Gleixner <tglx@...utronix.de>,
	linux-rt-users <linux-rt-users@...r.kernel.org>, akpm@...l.org,
	Ingo Molnar <mingo@...e.hu>,
	Clark Williams <clark.williams@...il.com>,
	Peter Zijlstra <peterz@...radead.org>,
	"Luis Claudio R. Goncalves" <lclaudio@...g.org>,
	Gregory Haskins <ghaskins@...ell.com>,
	Linus Torvalds <torvalds@...ux-foundation.org>,
	Andi Kleen <andi-suse@...stfloor.org>
Subject: Re: [PATCH][resubmit] x86: enable preemption in delay

On Sunday 01 June 2008 07:25:17 pm Andi Kleen wrote:
> >  static void delay_tsc(unsigned long loops)
> >  {
> > -	unsigned long bclock, now;
> > +	unsigned prev, prev_1, now;
> > +	unsigned left = loops;
> > +	unsigned prev_cpu, cpu;
> > +
> > +	preempt_disable();
> > +	rdtscl(prev);
> 
> 
> The unsigneds should be probably u64  and the rdtsc rdtscll.
> Otherwise this will all overflow for longer waits on a very
> fast systems (e.g. a 5Ghz system wraps 32bit in ~1.1 seconds)
> Normally such delays shouldn't be that long, but why risk
> overflow even in extreme cases?

Yes in principles, but the overflow (that could happen between
rdtscl(prev) and rdtscl(now) is taken into account the same way 
as in time_after()/time_before() macros, (differences only) see:

+		left -= now - prev;
.........
+	} while ((now-prev) < left);

If more than one overflow happen between rdtscl(prev) and 
rdtscl(now) (the task is suspended for a long time between two 
readings) all overflows after the first one will be lost. But the 
patch was submitted to guaranty minimum udelay() initially.
Sure, I could change to u64 if we reach a concensus.

Best regards.

Marin Mitov

> 
> -Andi
> 
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