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Date: Fri, 4 Dec 2009 11:21:19 +0200
From: "Ahmed S. Darwish" <darwish.07@gmail.com>
To: x86@kernel.org
Cc: Rusty Russell <rusty@rustcorp.com.au>, Ingo Molnar <mingo@redhat.com>,
       "H. Peter Anvin" <hpa@zytor.com>, linux-kernel@vger.kernel.org
Subject: x86: Is 'volatile' necessary for readb/writeb and friends?
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Hi all,

x86 memory-mapped IO register accessors cast the memory mapped address
parameter to a one with the 'volatile' type qualifier. For example, here
is readb() after cpp processing

--> arch/x86/include/asm/io.h:

static inline unsigned char readb(const volatile void __iomem *addr) {
	unsigned char ret;
	asm volatile("movb %1, %0"
		     :"=q" (ret)
		     :"m" (*(volatile unsigned char __force *)addr)
		     :"memory");
        return ret;
}

I wonder if the volatile qualifiers in the parameter above and at the asm
statement operand were strictly necessary, or just added for extra safety.

AFAIK, the asm statement already functions as a compiler barrier, and the
compiler won't 'optimize' the statement away due to the 'asm volatile' part,
so shouldn't things be safe without those volatile qualifiers?

The only red-herring I found in the gcc manual was the fact that the
"volatile asm instruction can be moved relative to other code, including
across jump instructions."

I wonder if this was the reason a volatile-type data dependency was added
to the mov{b,w,l,q} asm statements; not to reorder the asm instruction
around non-memory-accessing instructions (we already have a barrier).

Thank you!

-- 
Darwish