[Bug libstdc++/107049] error: ‘std::__cxx11::basic_string<_CharT, _Traits, _Alloc>::operator __sv_type() const [with _CharT = char; _Traits = std::char_traits; _Alloc = std::allocator; __sv_type = std::basic_string_view]’ is inaccessible within this context

[redi at gcc dot gnu.org via Gcc-bugs]

https://gcc.gnu.org/bugzilla/show_bug.cgi?id=107049

--- Comment #3 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to Jonathan Wakely from comment #2)
>   template<class T>
>     using If_sv = enable_if_t<is_convertible<const T&, string_view>::value>;

It works if this is changed to:

  template<class T>
    using If_sv = enable_if_t<__is_convertible(const T&, string_view)>;

So the problem is that it fails an access check when the built-in is used
outside the class body. But that should be a SFINAE context.


Further reduced:

template<bool B>
struct bool_constant { static constexpr bool value = B; };

template<typename From, typename To>
struct is_convertible
: public bool_constant<__is_convertible(From, To)>
{ };

template<bool> struct enable_if { };
template<> struct enable_if<true> { using type = void; };
template<bool B> using enable_if_t = typename enable_if<B>::type;


class Private
{
  operator int() const;

public:
  void f(const Private&);

  template<class T>
    enable_if_t<is_convertible<T, int>::value>
    f(T)
    { }
};

int main()
{
  Private p;
  p.f(p);
}

sfinae.cc: In instantiation of 'struct is_convertible<Private, int>':
sfinae.cc:23:5:   required by substitution of 'template<class T>
enable_if_t<is_convertible<T, int>::value> Private::f(T) [with T = Private]'
sfinae.cc:30:6:   required from here
sfinae.cc:6:24: error: 'Private::operator int() const' is private within this
context
    6 | : public bool_constant<__is_convertible(From, To)>
      |                        ^~~~~~~~~~~~~~~~~~~~~~~~~~
sfinae.cc:16:3: note: declared private here
   16 |   operator int() const;
      |   ^~~~~~~~