c++ - 为什么 std::ranges::filter_view 对象必须是非常量才能查询其元素？

Question

#include <ranges>
#include <iostream>
#include <string_view>

using namespace std::literals;

int main()
{
    auto fn_is_l = [](auto const c) { return c == 'l'; };

    {
        auto v = "hello"sv | std::views::filter(fn_is_l);
        std::cout << *v.begin() << std::endl; // ok
    }

    {
        auto const v = "hello"sv | std::views::filter(fn_is_l);
        std::cout << *v.begin() << std::endl; // error
    }
}

见: https://godbolt.org/z/vovvT19a5

<source>:18:30: error: passing 'const std::ranges::filter_view<
                       std::basic_string_view<char>, main()::
                       <lambda(auto:15)> >' as 'this' argument discards
                       qualifiers [-fpermissive]
   18 |         std::cout << *v.begin() << std::endl; // error
      |                       ~~~~~~~^~
In file included from <source>:1:/include/c++/11.1.0/ranges:1307:7: 
     note: in call to 'constexpr std::ranges::filter_view<_Vp, 
           _Pred>::_Iterator std::ranges::filter_view<_Vp, Pred>
           ::begin() [with _Vp = std::basic_string_view<char>; _Pred =
           main()::<lambda(auto:15)>]'
 1307 |       begin()
      |       ^~~~~

为什么一定要 std::ranges::filter_view 对象是查询其元素的非常量吗？

Answer 1

为了提供range所需的摊销常数时间复杂度, filter_view::begin将结果缓存在 *this 中.这会修改 *this 的内部状态因此不能在 const 中完成成员函数。