Re: C++ aliasing rules

[Jason Merrill <jason at redhat dot com>]

>>>>> "Dan" == Dan Nicolaescu <dann@godzilla.ICS.UCI.EDU> writes:

> mike stump <mrs@windriver.com> writes:

>> Consider:
>> 
>> struct { char buf[sizeof (double)]; short i; } secondo, *second = &secondo;
>> struct { double other; short i; } firsto, *not_really_first = &secondo;
>> 
>> and the question of whether or not first->i can ever alias second->i.
>> 
>> I suspect the closest we can come would be to
>> 
>> memcpy (&secondo, &firsto, sizeof (secondo));
>> 
>> and then play with not_really_first->i and second->i.
>> 
>> I think the answer is no, as no matter what you do, you need to
>> violate the notion of which struct type the object really was and you
>> cannot get at the short without going though the struct first, and in
>> the end, you cannot get through both struct types simultaneously.

> I like this answer. :-) 

> Does anybody disagree with Mike's statement above (wrt C++) ? Jason? 

I don't see how Mike's statement answers my question.  Given

 struct A { char buf[sizeof (double)]; short i; };
 struct B { double other; short i; };

 A a;
 A* ap = &a;
 B* bp = reinterpret_cast<B*>(ap);

referring to bp->i is accessing a.i through an lvalue of a class type (B)
which contains a member of a.i's type (short).  The standard seems to say
that's OK.

Again, if instead we consider a.i to be an indivisible part of 'a' for
purposes of [basic.lval], then the access would be undefined.  I would
probably support writing that into the standard.  But I don't think that's
what it says now.

Jason