Re: Is there a portable way to use pointer arithmetics with members.

From:
=?windows-1252?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sun, 28 Nov 2010 08:40:11 CST
Message-ID:
<ics0fb$7rq$1@news.eternal-september.org>
Am 27.11.2010 20:15, schrieb Elias Salom?o Helou Neto:

Hi,

Suppose I have the following class template:

template< unsigned n, class T>
class npoint {

     T head_;
     npoint< n - 1, T> tail_;

     public:

        T& get( unsigned i ) // Works with GCC 4.5
         {
            return( *(&head_ + i ) );
         }
};

template< class T>
class npoint< 0u, T> {};

My question is whether the get() member function is guaranteed to
work.


This is not guaranteed to work. The standard only guarantees for some special types (standard-layout structs) that a pointer to the structure can be reinterpreted as the pointer to the first initial member and vice versa. But 9.2/21 emphasizes in a note that this rule does not extend to the second or further members:

"[ Note: There might therefore be unnamed padding within a standard-layout struct object, but not at its beginning, as necessary to achieve appropriate alignment. ?end note ]"

If not, is there a portable way of accessing elements of an
object of such a class?


Yes, by making get a recursive function:

template< unsigned n, class T >
class npoint {
    template<unsigned, class> friend class npoint;
    T head_;
    npoint< n - 1, T > tail_;

    public:

       T& get(unsigned i)
       {
         return i == 0 ? head_ : tail_.get(i - 1);
       }
};

template< class T >
class npoint<0, T> {
    template<unsigned, class> friend class npoint;
    T& get( unsigned i ) {
      ... // Some error situation
    }
};

HTH & Greetings from Bremen,

Daniel Kr?gler

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