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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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