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> & );
>
>
>
>
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