I'm not sure this follows. It's stated explicitly that "The expression is evaluated and its value is discarded." How can you evaluate the expression without reading the volatle?
I'm certainly not an expert on this material but I wouldn't think you'd normally read a variable in order to evaluate it as an lvalue. My guess is that evaluating the expression means computing the address for something like "*(p+1)".
Also, there's the non-normative [Note: volatile is a hint to the implementation to avoid aggressive optimization involving the object because the value of the object might be changed by means undetectable by an implementation. See 1.9 for detailed semantics. In general, the semantics of volatile are intended to be the same in C++ as they are in C. ]
Yep.
Besides, all that changing this would do is break programs.
Well they're broken anyway since widely-used C++ compilers are choosing to not read the variable under these circumstances. The obvious solution is "don't do that" but there's plenty of code doing this out there, and plenty of developers who need re-educating.
John
