TITLE: evaluating &*e

(Newsgroups: comp.std.c++, 4 Jan 99)

FVALI: fvali@biotrack.com

>> In C (my reference is the C9x draft) the expression &*E  is equivalent
>> to E even if E is a null pointer, i.e., neither the operand of *, nor of
>> & is evaluated.
>>
>> I found no evidence that this expression is well-defined in Standard C++.
>>
>> Thus the following fragment is well defined C code but not C++:
>>
>> struct A { };
>> 
>> struct A* ap1 = 0;
>> struct A* ap2 = &*(ap1);
>> // assert( ap2 == 0 );
>> 
>> Why is this so?  What is the harm in defining this behavior as is done
>> for C (C9x atleast, I do not have access to a C89 std).


From: Steve Clamage <stephen.clamage@sun.com>

> Allowing &*p to have meaning when p is null requires a special
> case in the semantics and associativity of the operators. Without
> the special case, the meaning is &(*p), which is undefined for
> a null pointer.
>
> In C, you can add the language wart that the "&" and "*" cancel
> each other out, in effect mandating a compiler optimization.
>
> In C++, the situation is not so straightforward. Either or both of
> the operators might be overloaded. If so, the functions must
> be called unless the compiler can prove they have no side effects.
>
> As a further note, there is no special case in C89, which was
> the only C standard in effect when the C++ standard was under
> development. For that matter, C89 still is the only C standard
> in effect, since C9X is only in the CD1 stage.