Nope, that's not enough, it's not that simple. Yes your
example of a-a is of course straightforward but what about
b = (a & 1) | (b & 1);
if a is volatile and b is known to be odd, can the read of a be eliminated?
The answer should be no (and I think the standard guarantees this), but the
reasoning is completely different from thinking about the fact that a may
change unexpectedly, since obviously no matter what value comes from
reading a, b will be set to 1 if b is known to be odd.
Of course you can provide a pragmatic justification, think of a as a hardware
counter that counts the number of times it is referenced, then this should
count as a reference, even though on an as-if basis b would have the same
value.
The Ada standard says it clearly: a read or write of a volatile variable
is an external effect. period.
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