[語法] friend template operator overloading

[new052 (洞伍兩)]

==== 編譯環境 ====
Using built-in specs.
Target: i386-undermydesk-freebsd
Configured with: FreeBSD/i386 system compiler
Thread model: posix
gcc version 4.2.1 20070719  [FreeBSD]

==== 程式碼 ====
#include <iostream>

template <typename T>
class Test;

template <typename T>
std::ostream& operator<<(std::ostream&, const Test<T>&);

template <typename T>
class Test
{
private:
  T value;
public:
  Test(T value)
  {
    this->value = value;
  }
  Test()
  {
  }
  const Test<T> operator<<(int shift_bit) const;
  friend std::ostream& operator<< <T> (std::ostream&, const Test<T>&);
};

template <typename T>
const Test<T> Test<T>::operator<<(int shift_bit) const
{
  return *this;
}

template <typename T>
std::ostream& operator<<(std::ostream& out, const Test<T>& t)
{
  out << t.value;
  return out;
}

int main()
{
  Test<int> v(1234);
  std::cout << v << std::endl;
  return 0;
}
==== g++ -Wall 編譯訊息 ===
Test.cpp:23: error: declaration of 'operator<<' as non-function
Test.cpp:23: error: expected ';' before '<' token
Test.cpp: In function 'std::ostream& operator<<(std::ostream&, const Test<T>&) [with T = int]':
Test.cpp:43:   instantiated from here
Test.cpp:13: error: 'int Test<int>::value' is private
Test.cpp:36: error: within this context

==== 問題 ====
以上的程式碼沒有辦法編譯,
但如果把這兩行

  const Test<T> operator<<(int shift_bit) const;
  friend std::ostream& operator<< <T> (std::ostream&, const Test<T>&);

交換一下順序變成

  friend std::ostream& operator<< <T> (std::ostream&, const Test<T>&);
  const Test<T> operator<<(int shift_bit) const;

就可以編譯得過

想請問一下有經驗的前輩們,
這是 C++ 語法中有規定不能像我原本那種寫法,
還是只是編譯器的問題?

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