So Edward, how does now the paragraph #1 of the section «13.7.4 Friends» looks?
 
 
You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or http://ru.stackoverflow.com
 
  
>Четверг, 22 апреля 2021, 20:22 +03:00 от Edward Catmur <ecatmur_at_[hidden]>:
>
>This was all changed substantially by P1787R6: Declarations and where to find them[1]. Since then, using <T> or <> to specify that the friend in question is a template specialization is clearly necessary.
>
>It's definitely an MSVC bug that it continues to reject after making that amendment, though. I responded[2] to your SO question with more background suited to a general audience.
>
>1. http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2020/p1787r6.html
>2. https://stackoverflow.com/a/67217902/567292
>On Thu, 22 Apr 2021 at 16:38, Vladimir Grigoriev via Std-Discussion < std-discussion_at_[hidden] > wrote:
>>Bo, by the way I asked the question at Stackoverflow
>>
>>https://stackoverflow.com/questions/67203187/friend-template-operator?noredirect=1#comment118808492_67203187
>>
>>And it is interesting to note that at once it was down voted and closed because most of readers of the question thought that the friend declaration is not a template declaration and as a result they concluded that the question is a duplicate question.:)
>>
>>I was to reopen the question.
>>
>>
>>You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or http://ru.stackoverflow.com
>>
>>
>>>Четверг, 22 апреля 2021, 9:34 +03:00 от Bo Persson via Std-Discussion < std-discussion_at_[hidden] >:
>>>
>>>On 2021-04-21 at 23:21, Vladimir Grigoriev via Std-Discussion wrote:
>>>> The following program does not compile in MS Visual Studio 19.
>>>>
>>>> |#include <iostream> #include <string> template <typename T> class A;
>>>> template <typename T> std::ostream &operator <<( std::ostream &, const
>>>> A<T> & ); template <typename T> class A { private: T x; public: A( const
>>>> T &x ) : x( x ) {} friend std::ostream &::operator <<( std::ostream &,
>>>> const A<T> & ); }; template <typename T> std::ostream &operator <<(
>>>> std::ostream &os, const A<T> &a ) { return os << "a.x = " << a.x; } int
>>>> main() { std::cout << A<std::string>( "Hello" ) << '\n'; } |
>>>>
>>>> The compiler says that operator << is not a function.
>>>>
>>>> While the following program
>>>>
>>>> |#include <iostream> #include <string> template <typename T> class A;
>>>> template <typename T> std::ostream &f( std::ostream &, const A<T> & );
>>>> template <typename T> class A { private: T x; public: A( const T &x ) :
>>>> x( x ) {} friend std::ostream &::f( std::ostream &, const A<T> & ); };
>>>> template <typename T> std::ostream &f( std::ostream &os, const A<T> &a )
>>>> { return os << "a.x = " << a.x; } int main() { f( std::cout,
>>>> A<std::string>( "Hello" ) ) << '\n'; } |
>>>>
>>>> compiles successfully.
>>>>
>>>> What is the reason of that the first program does not compile? Is it a
>>>> bug of MS Visual Studio 19 or do I have missed something from the C++ 20
>>>> Standard?
>>>>
>>>> You can meet me at http://cpp.forum24.ru/ or www.stackoverflow.com or
>>>> http://ru.stackoverflow.com
>>>>
>>>I don't know the exact rule, but note that the friend is not a template,
>>>while the global operator is.
>>>
>>>MSVC accepts the code if you change it to
>>>
>>> template<typename U>
>>> friend std::ostream &::operator <<( std::ostream &, const A<U> & );
>>>
>>>
>>>
>>>
>>>--
>>>Std-Discussion mailing list
>>>Std-Discussion_at_[hidden]
>>>https://lists.isocpp.org/mailman/listinfo.cgi/std-discussion
>> --
>>Std-Discussion mailing list
>>Std-Discussion_at_[hidden]
>>https://lists.isocpp.org/mailman/listinfo.cgi/std-discussion