C++ Logo

STD-DISCUSSION

Advanced search

Subject: Friend template operator declaration.
From: Vladimir Grigoriev (vlad.moscow_at_[hidden])
Date: 2021-04-21 16:21:28


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



STD-DISCUSSION list run by std-discussion-owner@lists.isocpp.org

Older Archives on Google Groups