Re: Volatile MEMs in statement expressions and functions inlinedastrees

[Jason Merrill <jason at redhat dot com>]

>>>>> "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?

> 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