Clang报推导返回类型函数未定义错误，GCC可编译，谁符合C++23标准？

以下C++23代码可在GCC中正常编译并输出预期结果，但在Clang中编译失败：

#include <array>
#include <tuple>

constexpr auto makeNumberGenerator = [](this auto maker, int startingValue)
{
    return [maker, startingValue]<typename Generator>(this Generator generator) -> std::tuple<int, Generator>
    {
        return std::make_tuple(startingValue, maker(startingValue + 1));
    };
};

auto test = []{
    std::array<int, 5> arr;
    
    auto fn = [](this auto self, int* output, const int* end, auto callable) -> void
    {
        auto [value, newCallable] = callable();
        if (output != end)
        {
            *output = value;
            self(output + 1, end, newCallable);
        }
    };
    
    fn(std::begin(arr), std::end(arr), makeNumberGenerator(0));

    return arr;
}();

编译结果 #

编译参数为-O2 -std=c++23：

• GCC编译输出：

  test:
          .long   0
          .long   1
          .long   2
          .long   3
          .long   4
  

• Clang编译错误：

  <source>:9:47: error: function 'operator()<(lambda at <source>:5:38)>' with deduced return type cannot be used before it is defined
      9 |         return std::make_tuple(startingValue, maker(startingValue + 1));
        |                                               ^
  <source>:8:5: note: while substituting into a lambda expression here
      8 |     {
        |     ^
  <source>:26:59: note: in instantiation of function template specialization '(anonymous class)::operator()<(lambda at <source>:5:38)>' requested here
   26 |     fn(std::begin(arr), std::end(arr), makeNumberGenerator(0));
        |                                                           ^
  <source>:5:38: note: 'operator()<(lambda at <source>:5:38)>' declared here
    5 | constexpr auto makeNumberGenerator = [](this auto maker, int startingValue)
        |                                      ^
  1 error generated.
  Compiler returned: 1
  

核心疑问 #

该代码是否符合C++23标准？GCC与Clang哪个的行为是正确的？

编辑补充 #

编辑1：修复后可兼容两款编译器的代码 #

修改如下代码后，代码可在GCC和Clang中均编译通过：

constexpr auto makeNumberGenerator = [](this auto self, int startingValue)
{
    return std::bind([startingValue](auto maker)
    {
        return std::make_tuple(startingValue, maker(startingValue + 1));
    }, self);
};

若将std::bind替换为普通Lambda，仍会触发原错误。

编辑2：问题与deducing this特性无关 #

以下未使用deducing this特性的代码，同样出现GCC可编译、Clang报错的情况：

#include <array>
#include <tuple>

constexpr auto makeNumberGenerator = [](auto maker, int startingValue)
{
    return [maker, startingValue]<typename Generator>(Generator generator) -> std::tuple<int, Generator>
    {
        return std::make_tuple(startingValue, maker(maker, startingValue + 1));
    };
};

auto test = []{
    std::array<int, 5> arr;
    
    auto fn = [](auto self, int* output, const int* end, auto callable) -> void
    {
        auto [value, newCallable] = callable(callable);
        if (output != end)
        {
            *output = value;
            self(self, output + 1, end, newCallable);
        }
    };
    
    fn(fn, std::begin(arr), std::end(arr), makeNumberGenerator(makeNumberGenerator, 0));

    return arr;
}();

内容的提问来源于stack exchange，提问作者LHLaurini