Re: Why no std::back_insert_iterator::value_type?

From:
Howard Hinnant <howard.hinnant@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
22 Nov 2006 18:56:09 -0500
Message-ID:
<howard.hinnant-BBDDCE.12010822112006@syrcnyrdrs-02-ge0.nyroc.rr.com>
In article <4qV8h.51940$uv5.351593@twister1.libero.it>,
 Alberto Ganesh Barbati <AlbertoBarbati@libero.it> wrote:

Terry G ha scritto:

Because back_inserter_iterator<> is an output iterator but not an input
iterator. value_type, which is defined as the type able to hold the
value of the expression *it, makes sense only for input iterators.


What if I want to know what type *it will return?


The requirement of an output iterator says that the expression "*it = v"
is well-defined and does what it's supposed to do, but it's not said
what the expression "*it" is required to be and even if it can be used
in any way different from "*it = v". Therefore, defining value_type to
be void is a big warning sign that you shouldn't rely on the actual
type. It's like this for your own good. Why do you want to know?

Ganesh

PS: once we have decltype in the language, we could write decltype(*it),
but the question remains: what for?


I know why *I* want output iterators to have value_type. Because I want
to write generic code like:

template <class It>
It
format(char val, It first, It last)
{
    typedef typename std::iterator_traits<It>::value_type value_type;
    if (first == last)
        throw format_error();
    *first = static_cast<value_type>(val);
    return ++first;
}

I.e. I don't want the return type of *it (which should be it::reference
anyway, and it is fine with me if *it returns void). I want a type such
that if I convert my value v to that type, I'm assured of a clean
assignment into the output iterator.

Note that I'm also wanting output iterators to be equality comparable.
In the case of back_insert_iterator, the following definitions would be
fine:

bool operator==(const back_insert_iterator&,
                const back_insert_iterator&)
    {return false;}

bool operator!=(const back_insert_iterator&,
                const back_insert_iterator&)
    {return true;}

With such changes I could safely format a char into any "iterator
delimited stream".

template <class Container>
class back_insert_iterator
{
protected:
    Container* container;

public:
    typedef Container container_type;
    typedef output_iterator_tag iterator_category;
    typedef typename container_type::value_type value_type;
    typedef ptrdiff_t difference_type;
    typedef typename container_type::pointer pointer;
    typedef back_insert_iterator& reference;

    explicit back_insert_iterator(Container& x) : container(&x) {}
    back_insert_iterator& operator=(const value_type& value)
        {container->push_back(value); return *this;}
    back_insert_iterator& operator=(value_type&& value)
        {container->push_back(std::move(value)); return *this;}

    reference operator*() {return *this;}
    reference operator++() {return *this;}
    reference operator++(int) {return *this;}

    friend bool operator==(const back_insert_iterator&,
                           const back_insert_iterator&) {return false;}
    friend bool operator!=(const back_insert_iterator&,
                           const back_insert_iterator&) {return true;}
};

-Howard

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