Comparing the codes between compilers:
For the first code:

gcc compiles with a warning:
<source>:19:70: warning: friend declaration 'std::ostream& operator<<(std::ostream&, const A<T>&)' declares a non-template function [-Wnon-template-friend]
   19 |     friend std::ostream &::operator <<( std::ostream &, const A<T> & );
 
clang compiles without warnings.

icc compiles with two warnings:
<source>(19): warning #1098: the qualifier on this friend declaration is ignored
      friend std::ostream &::operator <<( std::ostream &, const A<T> & );
                           ^
<source>(19): warning #1624: "std::ostream &operator<<(std::ostream &, const A<T> &)" declares a non-template function -- add <> to refer to a template instance
      friend std::ostream &::operator <<( std::ostream &, const A<T> & );

And only msvc rejects the code:
<source>(19): error C2063: 'operator <<': not a function


For the second code, only msvc's behavior changes, and it now compiles without warnings.

(All tested on trunk/latest version available on godbolt, for x86)


On Thu, Apr 22, 2021 at 9:49 AM Vladimir Grigoriev via Std-Discussion <std-discussion@lists.isocpp.org> wrote:
Bo, I provided a reference that says about deduced template specialization. It is the case. That is if in a friend declaration there is used a qualified name (not a template-id) and a corresponding template declaration is found then the friend declaration is a templated specialization.
 
 
 
 
Четверг, 22 апреля 2021, 9:34 +03:00 от Bo Persson via Std-Discussion <std-discussion@lists.isocpp.org>:
 
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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