Re: value_type of an output iterator

On 30 Jul., 21:50, ShaunJ <sjack...@gmail.com> wrote:


Let me try to sketch the situation in form of a
dialogue:

You: "Just gimme something!"
OutputStream: "What do you want?"
You: "It doesn't matter - just something!"
Outputstream: "Sorry, I don't know how to
answer this question"

Joke aside: A *general* (or "pure") output
iterator does not have a fixed value type.
If this is true, you cannot require the
output stream to provide you one.

There is another way of looking at the problem:
Why does the function have the name "make_pairs"?
I assume, the name has some meaning. So, either
you have a concrete idea of the meaning or you
don't.

a) If you have an idea, you should use it, i.e.
require some special type or factory function.

b) If you don't have any idea, why don't you
use a heterogeneous container that fits
everything?

For (b), there is no reason for value_type. You could
just use std::make_pair (or std::make_tuple) as follows:

 template <typename It0, typename It1, typename OutIt>
 OutIt make_pairs(It0 first0, It0 last0, It1 first1, OutIt out)
 {
         while (first0 != last0)
                 *out++ = std::make_pair(*first0++, *first1++);
         return out;
 }

Of-course this means that out contains now std::pair<T, U>,
where T and U are the value types of It0 and It1. If this
doesn't fit your purpose, you should either

- Specify the intermediate output type (An idea that you
excluded)
- Impose requirements on the type OutIt that allows to
retrieve the value type. You could for example define
a separate traits type, that can be specialized by used
code. If you want, you could partially specialize the
traits for std::ostream_iterator to deduce the type T
(see below).

What I would agree with is, that it seems to be a reasonable
suggestion that std::ostream_iterator *does* provide the type
for you. Since it does not yet, try the traits approach, e.g.

template<class OutIter>
struct output_iterator_traits {
  typedef typename OutIter::value_type value_type;
  // ...
};

template<class T, class charT, class traits>
struct output_iterator_traits<std::ostream_iterator<T, charT, traits>


  typedef T value_type;
  // ...
};

HTH & Greetings from Bremen,

Daniel Kr?gler

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