Re: x86: Is 'volatile' necessary for readb/writeb and friends?

["H. Peter Anvin" <hpa@zytor.com>]

On 12/04/2009 06:39 AM, Segher Boessenkool wrote:
>> 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.

Wrong.  "volatile" here is an assertion that it is safe to pass pointer
to a volatile object to this function.

>>     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?  :-)

Again, this comes from the prototype being volatile.

Either way, it works, it is guaranteed to be safe, and removing it can
only introduce bugs, not remove them.

	-hpa

-- 
H. Peter Anvin, Intel Open Source Technology Center
I work for Intel.  I don't speak on their behalf.