[Bug libstdc++/15910] can't compile self defined void distance(std::vector<T>, std::vector<T>)

["giovannibajo at libero dot it" <gcc-bugzilla at gcc dot gnu dot org>]

------- Additional Comments From giovannibajo at libero dot it  2004-08-03 13:51 -------
Subject: Re:  can't compile self defined void distance(std::vector<T>, std::vector<T>)

pcarlini at suse dot de wrote:

> First, the ongoing work on type traits + tr1 may simplify
> considerably this kind of refinements, exploting enable_if idioms.

OK. Meanwhile, so that it does not seem I am here just to do FUD, this is
another proto-patch to fix this bug. This solution seems almost perfect to me:


template <class Iter>
typedef typename lazy_enable_if<
    has_difference_type<iterator_traits<T> >,
    get_difference_type<iterator_traits<T> >
>::type
distance(Iter first, Iter last) { ... }


with the following obvious definitions for the helpers:

template <bool C, class T> struct lazy_enable_if {};
template <class T> struct lazy_enable_if<true,T> { typedef typename T::type
type };

template <class T> struct get_difference_type {
    typedef typename T::difference_type type;
};

template <class T> struct has_difference_type {
    typedef char yes[1];
    typedef char no[2];

    template <class T> yes test(typename T::difference_type* );
    template <class T> no test(...);

    enum { result = (sizeof(test<T>(0)) == sizeof(yes)) };
};



> Second, if we really want to go this way, we should do it
> consistently: don't you believe that most definitely *many* other
> places of the library could be amenable to this treatment?

I don't know how many off-hand. Many other functions don't return a type which
is *indirectly* invalid if T is not an iterator (indirectly = causes an
instantiation error, rather than being directly invalid and trigger SFINAE), so
I don't think it affects v3 that much.

Giovanni Bajo




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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=15910