why does g++ need both definitions of template class and its friend function?

Question

I want to write a friend function for a template class. In visual studio, I can ignore the pre-definition both. But in g++, it is mandatory. Why?

#include <iostream>
using namespace std;
// g++ needs, vs do not needs
template <class T>
class A;

template <class T>
ostream & operator<<(ostream & c, const A<T> & v);
//- end of g++ needs

template <class T>
class A {
    T _v;
public:
    A() {}
    A(T v) : _v(v) {}
    friend ostream & operator<<<T>(ostream & c, const A<T> & v);
};
template <class T>
ostream & operator<<(ostream & c, const A<T> & v) {
    c << v._v; return c;
}

Answer 1

Because

friend ostream & operator<<<T>(ostream & c, const A<T> & v);

is a specialization of

template <class T>
ostream & operator<<(ostream & c, const A<T> & v);

you need to declare that first and the

A<T>

part means you have to declare that too before the operator declaration

template <class T>
class A;

So VS is probably wrong as C++14

14.5.4 Friends [temp.friend] 

gives the example

template<class T> class task;
template<class T> task<T>* preempt(task<T>*);

template<class T> class task {
  friend void next_time();
  friend void process(task<T>*);
  friend task<T>* preempt<T>(task<T>*);
  template<class C> friend int func(C);

  friend class task<int>;
  template<class P> friend class frd;
};

Where your example fits the 3rd friend declaration.