是否可以将“range-v3”包含与c++“range”包含混合？

Question

我有权使用GCC 10，并用-std=c++20编译，需要诸如generate cache1，concat等视图，这些视图可能要到C++23才会发布。

我在下面写了一个简单的程序，使用range-v3输出'1 2 2 3 3‘。我尝试将range-v3视图和ranges包含的视图混合在一起，但没有成功。“ranges”库似乎希望视图继承自view_interface，所以我创建了一个wrap_view_t来帮助实现这一点(未显示)。要查看工作范围-v3代码，请注释该行：

//#define MIX_RANGES_WITH_RANGE_V3

代码如下：

#include <iostream>

#define MIX_RANGES_WITH_RANGE_V3

#ifdef  MIX_RANGES_WITH_RANGE_V3
#include <ranges>
namespace test { using namespace std::ranges; }
#else
#include <range/v3/view/all.hpp>
#include <range/v3/view/interface.hpp>
#include <range/v3/view/join.hpp>
#include <range/v3/view/take_while.hpp>
#include <range/v3/view/transform.hpp>
namespace test { using namespace ranges; }
#endif
#include <range/v3/view/generate.hpp>
#include <range/v3/view/cache1.hpp>


auto my_view()
{
  auto gen{ ranges::views::generate( [ n = 0 ]() mutable
  {
    return std::vector< int >( ++n, n );
  } )
    | ranges::views::cache1
  };

  return std::move( gen )
    | test::views::take_while( []( auto const & value ) { return value.size() < 4; } )
    | test::views::transform( []( auto const & value )
    { return test::views::all( value ); } );

}

int main()
{
  for( auto const value : my_view() | test::views::join )
  {
    std::cout <<value <<" ";
  }
  std::cout <<std::endl;

  return 0;
}

和编译器错误

test_rng/test_rng_simpler.C: In function ‘auto my_view()’:
test_rng/test_rng_simpler.C:28:3: error: no match for ‘operator|’ (operand types are ‘std::remove_reference<ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >&>::type’ {aka ‘ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >’} and ‘std::ranges::views::__adaptor::_RangeAdaptorClosure<std::ranges::views::__adaptor::_RangeAdaptor<_Callable>::operator()<{my_view()::<lambda(const auto:16&)>}>::<lambda(_Range&&)> >’)
   27 |  return std::move( gen )
      |         ~~~~~~~~~~~~~~~~
      |                  |
      |                  std::remove_reference<ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >&>::type {aka ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >}
   28 |   | test::views::take_while( []( auto const & value ) { return value.size() < 4; } )
      |   ^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
      |                            |
      |                            std::ranges::views::__adaptor::_RangeAdaptorClosure<std::ranges::views::__adaptor::_RangeAdaptor<_Callable>::operator()<{my_view()::<lambda(const auto:16&)>}>::<lambda(_Range&&)> >
In file included from test_rng/test_rng_simpler.C:6:
/opt/rh/devtoolset-10/root/usr/include/c++/10/ranges:1180:4: note: candidate: ‘template<class _Tp> constexpr auto std::ranges::views::__adaptor::operator|(const std::ranges::views::__adaptor::_RangeAdaptorClosure<_Callable>&, const std::ranges::views::__adaptor::_RangeAdaptorClosure<std::ranges::views::__adaptor::_RangeAdaptor<_Callable>::operator()<{my_view()::<lambda(const auto:16&)>}>::<lambda(_Range&&)> >&)’
 1180 |    operator|(const _RangeAdaptorClosure<_Tp>& __x,
      |    ^~~~~~~~
/opt/rh/devtoolset-10/root/usr/include/c++/10/ranges:1180:4: note:   template argument deduction/substitution failed:
test_rng/test_rng_simpler.C:28:84: note:   ‘std::remove_reference<ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >&>::type’ {aka ‘ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >’} is not derived from ‘const std::ranges::views::__adaptor::_RangeAdaptorClosure<_Callable>’
   28 |   | test::views::take_while( []( auto const & value ) { return value.size() < 4; } )
      |                                                                                    ^
In file included from test_rng/test_rng_simpler.C:6:
/opt/rh/devtoolset-10/root/usr/include/c++/10/ranges:1175:4: note: candidate: ‘constexpr auto std::ranges::views::__adaptor::operator|(_Range&&, const std::ranges::views::__adaptor::_RangeAdaptorClosure<_Callable>&) [with _Range = ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >; _Callable = std::ranges::views::__adaptor::_RangeAdaptor<_Callable>::operator()<{my_view()::<lambda(const auto:16&)>}>::<lambda(_Range&&)>]’
 1175 |    operator|(_Range&& __r, const _RangeAdaptorClosure& __o)
      |    ^~~~~~~~
/opt/rh/devtoolset-10/root/usr/include/c++/10/ranges:1175:4: note: constraints not satisfied
test_rng/test_rng_simpler.C: In instantiation of ‘constexpr auto std::ranges::views::__adaptor::operator|(_Range&&, const std::ranges::views::__adaptor::_RangeAdaptorClosure<_Callable>&) [with _Range = ranges::cache1_view<ranges::generate_view<my_view()::<lambda()> > >; _Callable = std::ranges::views::__adaptor::_RangeAdaptor<_Callable>::operator()<{my_view()::<lambda(const auto:16&)>}>::<lambda(_Range&&)>]’:
test_rng/test_rng_simpler.C:28:84:   required from here
/opt/rh/devtoolset-10/root/usr/include/c++/10/ranges:78:13:   required for the satisfaction of ‘viewable_range<_Range>’ [with _Range = ranges::cache1_view<ranges::generate_view<my_view::._anon_131> >]
/opt/rh/devtoolset-10/root/usr/include/c++/10/ranges:79:31: note: no operand of the disjunction is satisfied
   79 |       && (borrowed_range<_Tp> || view<remove_cvref_t<_Tp>>);
      |          ~~~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~~~~~~~

也许我应该坚持'range-v3‘，直到我需要的所有视图在’range‘中都可用，包括？

Answer 1

由于range-v3中的适配器不继承ranges::view_interface，这使得它们不是标准中的view。为了避免悬空的危险，该标准禁止将不对borrowed_range或view (在您的示例中为gen)建模的range传递到管道中。

您需要做的是使用一个临时变量来接受generate，使其成为一个左值range，这样就可以满足borrowed_range并使用标准管道操作：

#include <iostream>
#include <ranges>
#include <range/v3/view/generate.hpp>
#include <range/v3/view/cache1.hpp>

int main() {
  auto gen = ranges::views::generate([n = 0]() mutable {
      return std::vector<int>(++n, n);
    }) | ranges::views::cache1;

  for (auto const value : gen
    | std::views::take_while([](auto const& value) { return value.size() < 4; })
    | std::views::transform([](auto const& value) { return std::views::all(value); })
    | std::views::join) {
      std::cout <<value <<" ";
  }
}

Demo.