Re: Volatile MEMs in statement expressions and functions inlinedastrees

[Paul Schlie <schlie at mediaone dot net>]

(please see annotations below)

>>>>>> "Paul" == Paul Schlie <schlie@mediaone.net> writes:
> 
>>>>>>>> "Paul" == Paul Schlie <schlie@mediaone.net> writes:
>>> 
>>>> (tt)(lv = rv) :: (lv = ((lt)tmp = rv, (lt)tmp), (tt)((lt)tmp))
>>> 
>>> Nope.  If this was the case, then (x += y) += z would add z to some random
>>> temporary.
>>> 
> 
>> Would have expected it to unambiguously be equivalent to:
> 
>>  x = x + y, x = x + z;
> 
> As would I.  Actually, I should have said '(x = y) += z'; based on your
> transformation above, I would expect this to become
> 
> tmp = y, x = tmp, tmp = tmp + z
> 
> whereas it should be
> 
> x = y, x = x + z
> 
> or am I misinterpreting you?

In which case I would re-write it as:

 x = y, x = x + z

 :: x = (x = ((xt)tmp = y, (xt)tmp)), ((xt)tmp2 = x, (xt)tmp2)
                                    + ((xt)tmp3 = y, (xt)tmp3))

-or-

 x = y + z

 :: x = ((xt)tmp = y, (xt)tmp) + ((xt)temp2 = z, (xt)tmp2)

Depending on adherence to an abstract rule requiring correlation between
the number and sequence of explicit original coded references to a volatile
within an expression, to it's correspondingly more restrictive semantic
interpretation.

> 
>> But in no case should the interpretation be ambiguous, some uniform rule
>> should be established.
> 
> Agreed.
> 
>>>> And it is agreed that:
>>> 
>>>> volatile q
>>>> (q = 10) == 10
>>> 
>>> No, that's where I disagree.
> 
>> It it being asserted that:
> 
>>  volatile q
>>  (q = 10) == 10
> 
>> is ambiguous, and therefore defined?
> 
> I am asserting that in C++ the above is unambiguously equivalent to
> 
> q = 10, q == 10
> 
> and therefore the result of the comparison cannot be predicted.
> 
> Jason

Hence you are asserting that "(q = 10) == 10" may not be true, if q is
volatile. Correct?

-paul-