Re: [PATCHv3 1/4] x86: provide xadd()

[David Vrabel <david.vrabel@citrix.com>]

On 21/04/15 14:12, Jan Beulich wrote:
>>>> On 21.04.15 at 14:36, <david.vrabel@citrix.com> wrote:
>> On 21/04/15 11:36, Jan Beulich wrote:
>>>>>> On 21.04.15 at 12:11, <david.vrabel@citrix.com> wrote:
>>>> +static always_inline unsigned long __xadd(
>>>> +    volatile void *ptr, unsigned long v, int size)
>>>> +{
>>>> +    switch ( size )
>>>> +    {
>>>> +    case 1:
>>>> +        asm volatile ( "lock; xaddb %b0,%1"
>>>> +                       : "+r" (v), "+m" (*__xg((volatile void *)ptr))
>>>> +                       :: "memory");
>>>> +        return v;
>>>
>>> This doesn't seem to guarantee to return the old value: When the
>>> passed in v has more than 8 significant bits (which will get ignored
>>> as input), nothing will zap those bits from the register. Same for
>>> the 16-bit case obviously.
>>>
>>>> +#define xadd(ptr, v) ({                                         \
>>>> +            __xadd((ptr), (unsigned long)(v), sizeof(*(ptr)));  \
>>>> +        })
>>>
>>> Assuming only xadd() is supposed to be used directly, perhaps
>>> the easiest would be to cast v to typeof(*(ptr)) (instead of
>>> unsigned long) here?
>>
>> I don't see how this helps.  Did you perhaps mean cast the result?
>>
>> #define xadd(ptr, v) ({                                    \
>>             (typeof *(ptr))__xadd(ptr, (unsigned long)(v), \
>>                                   sizeof(*(ptr)));         \
>>         })
> 
> Casting the result would work too; casting the input would have
> the same effect because (as said) the actual xadd doesn't alter
> bits 8...63 (or 16...63 in the 16-bit case), i.e. whether zero
> extension happens before or after doing the xadd doesn't matter.

Oh yes, of course.  Any preference to which method?

David