如何将一对迭代器转换为视图？

如何解决如何将一对迭代器转换为视图？

我有一对迭代器，我想在它上面使用 ranges::views::filter(some_predicate)（使用管道运算符）。 AFAIU 我应该首先将我的一对迭代器转换为视图。我尝试使用 ranges::subrange(first,last) 来执行此操作，但收到了可怕的错误消息。

注1：我使用的是 C++14 和 range-v3 版本 0.9.1（与 gcc-5.5 兼容的最后一个版本）。如果使用 C++17/20 和/或使用 C++20 std::ranges 时解决方案不同，我也很想知道发生了什么变化。

注 2：我发现 range-v3 的 documentation 严重缺乏，所以我使用 cppreference.com。如果你知道更好的文档，我很感兴趣。

编辑：

在我的真实代码中，我包装了一个 java 风格的遗留迭代器（它有一个 next() 方法而不是 operator++/operator*。我将它们包装在 C++ 中-兼容的包装器。然后我尝试将该包装器转换为视图，最后对其进行过滤。我在 godbolt 上重现了一个最小示例。按照建议使用 iterator_range，但它仍然无法编译（请参阅下面的第二个编辑）。

#include "range/v3/all.hpp"
#include "range/v3/iterator_range.hpp"

class LegacyIteratorWrapper {
public:
    using value_type = int;
    using difference_type = std::ptrdiff_t;
    using pointer = value_type*;
    using reference = value_type&;
    using iterator_category = std::input_iterator_tag;

    // The type isn’t default-constructible,the error comes from here
    LegacyIteratorWrapper() = delete;

    static LegacyIteratorWrapper create();
    
    reference operator*() const;
    pointer operator->();
    LegacyIteratorWrapper& operator++();
    LegacyIteratorWrapper operator++(int);
    friend bool operator==(const LegacyIteratorWrapper& a,const LegacyIteratorWrapper& b);
    friend bool operator!=(const LegacyIteratorWrapper& a,const LegacyIteratorWrapper& b);
};

void foo()
{
    LegacyIteratorWrapper begin { LegacyIteratorWrapper::create() };
    LegacyIteratorWrapper end { LegacyIteratorWrapper::create() };
    ranges::iterator_range<LegacyIteratorWrapper,LegacyIteratorWrapper> rng {begin,end};
    auto _ = rng
        | ranges::views::filter(
            [&](auto _) { return true; }
          )
        ;
}

In file included from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/functional/reference_wrapper.hpp:24:0,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/detail/variant.hpp:33,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/iterator/common_iterator.hpp:26,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/view/interface.hpp:24,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/view/ref.hpp:25,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/action/action.hpp:29,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/action.hpp:17,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/all.hpp:17,from <source>:1:
/opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/functional/pipeable.hpp: In instantiation of 'constexpr auto ranges::operator|(Arg&&,Pipe) [with Arg = ranges::iterator_range<LegacyIteratorWrapper,LegacyIteratorWrapper>&; Pipe = ranges::views::view<ranges::make_pipeable_fn::operator()(Fun) const [with Fun = ranges::detail::bind_back_fn_<ranges::views::cpp20_filter_fn,foo()::<lambda(auto:15)> >]::_>; bool CPP_false_ = false; typename concepts::detail::identity<typename std::enable_if<(static_cast<bool>(((! is_pipeable_v<Arg>) && is_pipeable_v<Pipe>)) || CPP_false_),void>::type>::invoke<int> <anonymous> = 0]':
<source>:33:11:   required from here
/opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/functional/pipeable.hpp:63:53: error: no matching function for call to 'ranges::pipeable_access::impl<ranges::views::view<ranges::make_pipeable_fn::operator()(Fun) const [with Fun = ranges::detail::bind_back_fn_<ranges::views::cpp20_filter_fn,foo()::<lambda(auto:15)> >]::_> >::pipe(ranges::iterator_range<LegacyIteratorWrapper,LegacyIteratorWrapper>&,ranges::views::view<ranges::make_pipeable_fn::operator()(Fun) const [with Fun = ranges::detail::bind_back_fn_<ranges::views::cpp20_filter_fn,foo()::<lambda(auto:15)> >]::_>&)'
             return pipeable_access::impl<Pipe>::pipe(static_cast<Arg &&>(arg),pipe);
                                                     ^
In file included from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/range_fwd.hpp:22:0,from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/action/action.hpp:21,from <source>:1:
/opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/view/view.hpp:114:35: note: candidate: template<class Rng,class Vw> static constexpr auto ranges::views::view<View>::pipe(Rng&&,Vw&&,concepts::detail::enable_if_t<concepts::detail::Nil,(static_cast<bool>((viewable_range<Rng> && invocable<View&,Rng>)) || concepts::detail::CPP_false(concepts::detail::Nil{}))>) [with Rng = Rng; Vw = Vw; View = ranges::make_pipeable_fn::operator()(Fun) const [with Fun = ranges::detail::bind_back_fn_<ranges::views::cpp20_filter_fn,foo()::<lambda(auto:15)> >]::_]
             static constexpr auto CPP_fun(pipe)(Rng && rng,Vw && v)( //
                                   ^
/opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/view/view.hpp:114:35: note:   template argument deduction/substitution Failed:
<source>: In function 'void foo()':
<source>:33:11: error: 'void _' has incomplete type
           )
           ^
ASM generation compiler returned: 1
In file included from /opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/functional/reference_wrapper.hpp:24:0,Vw && v)( //
                                   ^
/opt/compiler-explorer/libs/rangesv3/0.9.1/include/range/v3/view/view.hpp:114:35: note:   template argument deduction/substitution Failed:
<source>: In function 'void foo()':
<source>:33:11: error: 'void _' has incomplete type
           )
           ^
Execution build compiler returned: 1

EDIT2（已解决）：我收到一个错误，因为 LegacyIteratorWrapper 不是默认可构造的。这是满足常规特征（由 range-v3 模拟）所必需的，这是 C++ 迭代器所必需的，并且包括可默认构造。

解决方法

由于您还没有真正给出代码来支持您得到可怕的错误消息，这里的代码在一个向量上定义了两个迭代器，利用了 ranges::subrange，并过滤了结果范围。

#include <iostream>
#include <range/v3/view/filter.hpp>
#include <range/v3/view/subrange.hpp>
#include <vector>

namespace r = ranges;
namespace rv = ranges::views;

int main() {
    std::vector<int> v{1,2,3,4,5};
    auto even = [](auto x){ return x % 2 == 0; };
    auto i1 = v.begin() + 1;
    auto ie = v.end() - 1;
    auto w = r::subrange(i1,ie) | rv::filter(even);
    for (auto i : w) {
        std::cout << i << std::endl;
    }
}

如果您阐明什么您的示例代码是什么以及您得到了什么错误，我或其他人可以给您一个更好的答案。

在 range-v3 中，有 iterator_range 可以用来将迭代器包装到一个范围对象中。

在 C++20 中，您可以使用 std::span 将这些迭代器包装到一个范围对象中

您可以使用以下方法将每对相等的迭代器转换为 std::ranges::subrange：

auto my_range = std::ranges::subrange{begi,endi};