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0 pointscperciva11y ago0 comments

The key text is in C99 §6.7.4: "An object that has volatile-qualified type...". Note that this does not say "An object accessed via a pointer to a volatile-qualified type": What matters is the type of the underlying object, i.e., how it is originally defined.

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cnvogel11y ago

Ok, that's quite some language-lawyering going on here ;-), and while I enjoy it, I'm completely aware of the fact that in practice the behavior of actual implementation (in software, compilers, operating systems) always trumps theoretic considerations.

That being said...

§6.5.3.1/4 The unary * operator denotes indirection. (...) the result is an lvalue designating the object.

So, I believe that there will be no difference in access to something declared as

    volatile X v;
    v = ...;

and

    X v;
    *((volatile X*)&v) = ...;

if the access to v will always be performed through such a cast, as ((volatile..)) will always denote the same "volatile" X object stored in v, independent of the fact that storage to the volatile X object is allocated in a "nonvolatile" X object.

cpercivaOP11y ago

In practice most compilers respect the volatile keyword on a pointer target type. But if you want to conform to the standard you need to look at what the standard says, not merely how it's commonly implemented.

    §6.5.3.1/4 The unary * operator denotes indirection.
    (...) the result is an lvalue designating the object.

Yes, but calling the sky green doesn't make it so. Dereferencing a pointer to an object yields that object, and the object is volatile or not depending on whether it was declared as volatile; whether you say that it's volatile before you dereference the pointer is irrelevant. (That is, irrelevant to the requirements of the standard; it's very relevant to most compilers, because it's usually impossible to prove whether the object ultimately being accessed is volatile or not.)

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