Why aren't ranges' algorithms compatible w

2020-02-09 08:57发布

问题:

#include <vector>
#include <iostream>
#include <range/v3/all.hpp>

int main()
{
    auto coll = std::vector{ 1, 2, 3 };

    ranges::copy(
        coll,
        ranges::ostream_iterator<int>{  std::cout, ", " }
    ); // ok

    ranges::copy(
        coll, 
        std::ostream_iterator<int>{ std::cout, ", " }
    ); // error 
}

The issue is shown in the code above. I use ranges-v3-0.3.7.

To me, the generic algorithm copy shouldn't care about the destination iterator type as long as it meets the requirements of output iterator.

If so, why aren't ranges' algorithms compatible with std's iterators?

回答1:

To me, the generic algorithm copy shouldn't care about the destination iterator type as long as it meets the requirements of output iterator.

This is correct. It's not that ranges::copy somehow recognizes ranges::ostream_iterator and not std::ostream_iterator. It's that Ranges has a refined concept for what an OutputIterator is, such that ranges::ostream_iterator does model OutputIterator but std::ostream_iterator does not.

Specifically, ranges::copy() requires WeaklyIncrementable<O> which refines SemiRegular<O> which requires DefaultConstructible. ranges::ostream_iterator is default constructible, but std::ostream_iterator is not.

Hence the failure.


In P0896, the range-based copy() algorithm does require WeaklyIncrementable (and thus DefaultConstructible) for its output iterator - but addresses this mismatch by also adding a default constructor to std::ostream_iterator (see page 70).


Note that the range-v3/Ranges TS/Ranges Proposal concept OutputIterator is separate from the standard library's existing concept of OutputIterator. std::ostream_iterator does not model the former but it does model the latter - so using std::copy with a std::ostream_iterator today is perfectly fine. And post-P0896, using ranges::copy with a std::ostream_iterator will also be fine - because of the proposed changes to std::ostream_iterator.