为什么我不能使用range-v3反转分段范围？

Question

我想拆分，反向，然后使用range-v3加入一个字符串。 但是，下面的代码将无法编译。

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

using namespace ranges;

int main(int argc, char *argv[])
{
    auto str = std::string("abc.def.ghi");
    auto sv = str
              | view::split('.')
              | view::reverse
              | view::join('.');
    std::cout<<sv;
    return 0;
}

编译器输出：

error: invalid operands to binary expression ('decltype(pipeable_access::impl<view<reverse_fn> >::pipe(static_cast<ranges::v3::split_view<ranges::v3::iterator_range<std::_String_iterator<std::_String_val<std::_Simple_types<char> > >, std::_String_iterator<std::_String_val<std::_Simple_types<char> > > >, ranges::v3::view::split_fn::element_pred<std::basic_string<char, std::char_traits<char>, std::allocator<char> > &> > &&>(arg), pipe))' (aka 'void') and 'decltype(make_view(view_access::impl<join_fn>::bind(this->view_, static_cast<char &&>(ts))))' (aka 'view<ranges::v3::detail::pipeable_binder<std::_Binder<std::_Unforced, ranges::v3::view::join_fn &, const std::_Ph<1> &, char> > >'))
range\v3\view\any_view.hpp:60: candidate function not viable: cannot convert argument of incomplete type 'decltype(pipeable_access::impl<view<reverse_fn> >::pipe(static_cast<ranges::v3::split_view<ranges::v3::iterator_range<std::_String_iterator<std::_String_val<std::_Simple_types<char> > >, std::_String_iterator<std::_String_val<std::_Simple_types<char> > > >, ranges::v3::view::split_fn::element_pred<std::basic_string<char, std::char_traits<char>, std::allocator<char> > &> > &&>(arg), pipe))' (aka 'void') to 'ranges::v3::category' for 1st argument
C:\Program Files (x86)\Microsoft Visual Studio\2019\Community\VC\Tools\MSVC\14.21.27702\include\regex:1219: candidate function not viable: cannot convert argument of incomplete type 'decltype(pipeable_access::impl<view<reverse_fn> >::pipe(static_cast<ranges::v3::split_view<ranges::v3::iterator_range<std::_String_iterator<std::_String_val<std::_Simple_types<char> > >, std::_String_iterator<std::_String_val<std::_Simple_types<char> > > >, ranges::v3::view::split_fn::element_pred<std::basic_string<char, std::char_traits<char>, std::allocator<char> > &> > &&>(arg), pipe))' (aka 'void') to 'std::_Node_flags' for 1st argument
range\v3\utility\functional.hpp:725: candidate template ignored: substitution failure [with Arg = void, Pipe = ranges::v3::view::view<ranges::v3::detail::pipeable_binder<std::_Binder<std::_Unforced, ranges::v3::view::join_fn &, const std::_Ph<1> &, char> > >, _concept_requires_724 = false, $3 = nullptr]: cannot form a reference to 'void'
range\v3\utility\functional.hpp:734: candidate template ignored: requirement 'false || (is_pipeable<void>() && is_pipeable<ranges::v3::view::view<ranges::v3::detail::pipeable_binder<std::_Binder<std::_Unforced, ranges::v3::view::join_fn &, const std::_Ph<1> &, char> > > >())' was not satisfied [with Pipe0 = void, Pipe1 = ranges::v3::view::view<ranges::v3::detail::pipeable_binder<std::_Binder<std::_Unforced, ranges::v3::view::join_fn &, const std::_Ph<1> &, char> > >, _concept_requires_733 = false]

找到了这样做的方法：

    auto temp_container = str
              | view::split('.')
              | ::ranges::to_vector
              | action::reverse
              ;
    std::string output = temp_container 
              | view::all
              | view::join('.')
              ;

有更好的想法吗？

Answer 1

view::split返回输入范围或前进范围，具体取决于适应的范围。 view::reverse至少需要一个双向范围才能向后迭代。

原则上，可以实现双向范围的split_view 。 问题是构造函数已经需要完全遍历适应范围以找到最后一个元素，使其成为o（n）操作。 据我所知，观点的构建必须是o（1）。