Re: [PATCH] Linux Kernel Markers 0.13 for 2.6.17

From: Mathieu Desnoyers
Date: Mon Sep 25 2006 - 20:59:46 EST

* Mathieu Desnoyers (mathieu.desnoyers@xxxxxxxxxx) wrote:
> Yes, preempt_disable() has a barrier(), on gcc :
> __asm__ __volatile__("": : :"memory").
> > Either way, this doesn't prevent some otherwise unrelated
> > non-memory-using code from being scheduled in there, which would not be
> > executed. The gcc manual really strongly discourages jumping between
> > inline asms, because they have basically unpredictable results.
> >
> Ok, I will do the call in assembly then.

Before I rush on a solution too fast... I have a question for you :

To protect code from being preempted, the macros preempt_disable and
preempt_enable must normally be used. Logically, this macro must make sure gcc
doesn't interleave preemptible code and non-preemptible code.

Starting with this hypothesis, what makes gcc aware of this ? If we check
preempt_disable (the disable call is almost symmetric) :

define add_preempt_count(val) do { preempt_count() += (val); } while (0)

#define inc_preempt_count() add_preempt_count(1)

#define preempt_disable() \
do { \
inc_preempt_count(); \
barrier(); \
} while (0)

So the magic must be in the barrier() macro :

/* Optimization barrier */
/* The "volatile" is due to gcc bugs */
#define barrier() __asm__ __volatile__("": : :"memory")

Which makes me think that if I put barriers around my asm, call, asm trio, no
other code will be interleaved. Is it right ?


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