Re: expression statements, volatiles, and C vs. C++

[John Regehr <regehr at cs dot utah dot edu>]

> 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