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Message-Id: <8B1FEF0C-5D71-4D64-ADC3-1EE60F50779F@kernel.crashing.org>
Date: Fri, 4 Dec 2009 15:39:06 +0100
From: Segher Boessenkool <segher@...nel.crashing.org>
To: "Ahmed S. Darwish" <darwish.07@...il.com>
Cc: x86@...nel.org, Rusty Russell <rusty@...tcorp.com.au>,
Ingo Molnar <mingo@...hat.com>,
"H. Peter Anvin" <hpa@...or.com>, linux-kernel@...r.kernel.org
Subject: Re: x86: Is 'volatile' necessary for readb/writeb and friends?
> 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) {
This "volatile" is meaningless.
> unsigned char ret;
> asm volatile("movb %1, %0"
This "volatile" is required; without it, if "ret" isn't used (or can
be optimised away), the asm() could be optimised away.
> :"=q" (ret)
> :"m" (*(volatile unsigned char __force *)addr)
This "volatile" has no effect, since the asm has a "memory" clobber.
Without that clobber, this "volatile" would prevent moving the asm
over other memory accesses.
If you want to get all language-lawyery, if the object pointed to by
"addr" is volatile, the volatile here _is_ needed: accessing volatile
objects via a not volatile-qualified lvalue is undefined. But since
this is GCC-specific code anyway, do you care? :-)
> :"memory");
> return ret;
> }
Segher
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