友元声明声明一个非模板函数，未定义的引用-解网

问：

我有这样的模板类，其中包括一个朋友方法：

template<class T1, class T2, int n>
class Graph final
{
private:
    std::array<T1, n> vertex;
    std::array<std::array<T2, n>, n> arcs;
public:
    Graph(const std::array<T1, n> & v, const std::array<std::array<T2, n>, n> & a);
    friend std::ostream & operator<<(std::ostream & os, const Graph<T1, T2, n> & g);
};

template<class T1, class T2, int n>
Graph<T1, T2, n>::Graph(const std::array<T1, n> & v, const std::array<std::array<T2, n>, n> & a)
    : vertex(v)
{ ... }

template<class T1, class T2, int n>
std::ostream & operator<<(std::ostream & os, const Graph<T1, T2, n> & g)
{ ... }

此类用于保存图形。Vertex 是存储节点的数组，T1 是对应的元素类型。Arcs 是一个 n * n 的二维数组，用于存储节点之间的路径开销，T2 是对应的元素类型。（请注意，n 是模板非类型参数。

我尝试创建一个像这样调用朋友的方法：Graph

void test() {
    using std::array;
    array<int, 5> vertex = {100, 200, 300, 400, 500};

    array<int, 5> a = {2, 3, 2, 4, 5};
    array<array<int, 5>, 5> arcs = {a, a, a, a, a};

    Graph<int, int, 5> g(vertex, arcs);
    std::cout << g << std::endl;
}

仅得到错误：

tree_mst.h:31:87: warning: friend declaration ‘std::ostream& meyok::operator<<(std::ostream&, const Graph<T1, T2, n>&)’ declares a non-template function [-Wnon-template-friend]
         friend std::ostream & operator<<(std::ostream & os, const Graph<T1, T2, n> & g);
                                                                                       ^
tree_mst.h:31:87: note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)

../lib/libmtree.so: undefined reference to `operator<<(std::ostream&, Graph<int, int, 5ull> const&)'
../lib/libmtree.so: undefined reference to `Graph<int, int, 5ull>::Graph(std::array<int, 5ul> const&, std::array<std::array<int, 5ul>, 5ul> const&)'

为什么这里会弹出警告和错误。我该如何解决？

C++11 未定义引用

// Forward declaration
template<class T1, class T2, int n>
class Graph;

// Forward declaration
template<class T1, class T2, int n>
std::ostream & operator<<(std::ostream & os, const Graph<T1, T2, n> & g);

template<class T1, class T2, int n>
class Graph
{
public:
    friend std::ostream & operator<< <T1, T2, n>(std::ostream & os, const Graph<T1, T2, n> & g);
};

但是，完全避免通常更简单，并将委托给成员函数：friendoperator<<

template<class T1, class T2, int n>
class Graph
{
public:
    void print(std::ostream& os) const { /* actual output logic here */);
};

template<class T1, class T2, int n>
std::ostream& operator<<(std::ostream& os, const Graph<T1, T2, n>& g) {
  g.print(os);
  return os;
}

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友元声明声明一个非模板函数，未定义的引用

friend declaration declares a non-template function, undefined reference

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