On Friday 15 November 2002 04:45 pm, Gennaro Prota wrote:
> Now, let me say that there's an explicit
> contradiction (as usual in such scenarios where a "special" case is assumed
> to be specified separately) between 5.2.10/10 and 5.2.10/7, because the
> latter says that in
>
> reinterpret_cast<T*>(&x)
>
> you can only cast back the result to the source type, whereas 5.2.10/10
> defines the reference-cast in terms of pointer-cast and says it's ok to
> *dereference* the resulting value.
I don't see the contradiction here. 5.2.10/7 says that you can cast from a T
pointer to a U pointer and back to a T pointer and get the original pointer
back. 5.2.10/10 says that reinterpret_cast does the same for references (by
specifying the semantics of reinterpret_cast on references in terms of
reinterpret_cast on pointers). The dereference (and address-of, for that
matter) operators are only there to perform the pointer/reference conversions
to that the semantic equivalent can be specified.
If there's a contradiction between these two paragraphs, then there has to be
some piece of code that can be interpreted in two different ways. Do we have
such an example?
> However the real issue is that we have
> first
>
> T -> cv char& -> char &
>
> and then
>
> char * -> T*
>
> The latter is a reinterpret_cast from an rvalue of type char*, so except
> that I can cast it back to a char* the result is unspecified.
By 5.2.10/10, the first part of that is semantically equivalent to:
T* -> cv char* ->char*
> In short, I agree with you that probably the intent is for it to work. But
> I think that should be clarified.
>
> Genny.
I've given probably the closest thing to a proof that one can get without
having a formal specification. It uses only semantic equivalences given in
the three paragraphs it cites. Is there a problem with it?
Doug
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