While I'm trying to understand the ostream_joiner's std::ranges algorithm requirement concept failure fiasco, I found that GCC implementation of WeaklyIncrementable iterator looks somewhat different from the standard.

GCC 10.2 stdlibc++ "iterator_concept.h" defines weakly_incrementable also requires

  • moveable
  • __is_signed_integer_like<iter_difference_t<_Iter>>
    which seems to requires Iter::different_type is signed integral or can be "long long".
   template<typename _Tp> requires requires { typename _Tp::difference_type; }
     struct incrementable_traits<_Tp>
     { using difference_type = typename _Tp::difference_type; };
  
   template<typename _Tp>
     requires (!requires { typename _Tp::difference_type; }
               && requires(const _Tp& __a, const _Tp& __b)
               {
                 requires (!is_void_v<remove_pointer_t<_Tp>>); // PR c++/78173
                 { __a - __b } -> integral;
               })
     struct incrementable_traits<_Tp>
     {
       using difference_type
         = make_signed_t<decltype(std::declval<_Tp>() - std::declval<_Tp>())>;
     };
   ...
   template<typename _Tp>
     using iter_difference_t = __detail::__iter_diff_t<remove_cvref_t<_Tp>>;
   ...
     using __max_diff_type = long long;
     ...
     template<typename _Tp>
       concept __is_signed_integer_like = signed_integral<_Tp>
         || same_as<_Tp, __max_diff_type>;
   ...
   /// Requirements on types that can be incremented with ++.
   template<typename _Iter>
     concept weakly_incrementable = default_initializable<_Iter>
       && movable<_Iter>
       && requires(_Iter __i)
       {
         typename iter_difference_t<_Iter>;
         requires __detail::__is_signed_integer_like<iter_difference_t<_Iter>>;
         { ++__i } -> same_as<_Iter&>;
         __i++;
       };

This seems more strict than semiregular in the standard that requires only std::copyable.

Why? What benefits do they give?

Is it okey/possible/should to write ostream_joiner that satisfies WeaklyIncrementable iterator?

0

There are 0 answers