问：

下面是一个涉及两个友元算子的 MRE 示例：和：operator<<operator-

#include <ostream>

template <typename T>
class container;
template <typename T>
std::ostream operator<<(std::ostream & os, const container<T> & data);
template <typename T>
container<T> operator-(const int & first, const container<T> & second);

template <typename T>
class container
{
    public:
        container<T> operator-(const container<T> & other);
        friend std::ostream operator<< <>(std::ostream & os, const container<T> & data); // this one works
        friend container<T> operator- <>(const int & first, const container<T> & second); // error: declaration of 'operator-' as non-function
};

运算符的实际定义不在这里（也不相关）。如注释所示，编译但结果为：operator<<operator-

g++：

<source>:16:29: error: declaration of 'operator-' as non-function
   16 |         friend container<T> operator- <>(const int & first, const container<T> & second); // error: declaration of 'operator-' as non-function
      |                             ^~~~~~~~
<source>:16:37: error: expected ';' at end of member declaration
   16 |         friend container<T> operator- <>(const int & first, const container<T> & second); // error: declaration of 'operator-' as non-function
      |                                     ^
      |                                      ;
<source>:16:39: error: expected unqualified-id before '<' token
   16 |         friend container<T> operator- <>(const int & first, const container<T> & second); // error: declaration of 'operator-' as non-function
      |                                       ^

咔嚓++：

<source>:16:29: error: friends can only be classes or functions
        friend container<T> operator- <>(const int & first, const container<T> & second); // error: declaration of 'operator-' as non-function
                            ^
<source>:16:38: error: expected ';' at end of declaration list
        friend container<T> operator- <>(const int & first, const container<T> & second); // error: declaration of 'operator-' as non-function
                                     ^
                                     ;
2 errors generated.

（非友元）已被声明的事实使该友元成为“非函数”。operator-operator-

为什么有效而无效？我该如何解决？朋友是否需要额外的模板参数（不是）？operator<<operator-operator-Toperator-

编辑

该问题与 c++17 密切相关。它显然在 c++20 上编译良好。

17 C ++20 模板好友

template <typename T>
class container;

template <typename T>
std::ostream& operator<<(std::ostream & os, const container<T> & data) { return os; }

template <typename T>
container<T> operator-(const int & first, const container<T> & second) {return second;}

template <typename T>
class container
{
    public:
        container<T> operator-(const container<T> & other);

        template <typename>
        friend std::ostream& operator<<(std::ostream & os, const container<T> & data);
        
        template <typename>
        friend container<T> operator- (const int & first, const container<T> & second);
};

现场演示：godbolt.org/z/Tbz1Yvf5T

编辑：我注意到，如果您更改顺序（先联系操作员）。它编译得很好。自c++11c++20

template <typename T>
class container;

template <typename T>
std::ostream operator<<(std::ostream & os, const container<T> & data);

template <typename T>
container<T> operator-(const int & first, const container<T> & second);

template <typename T>
class container
{
    public:
        friend std::ostream operator<< <>(std::ostream & os, const container<T> & data); // this one works
        friend container<T> operator- <>(const int & first, const container<T> & second); // error: declaration of 'operator-' as non-function

        // now it compiles
        container<T> operator-(const container<T> & other);
};

定义模板类的友元运算符时不一致

Inconsistency when defining friend operators of template class

编辑

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