基于模板参数的友元重载opearto<<常见问题

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对于模板下的类定义，一般重载operator<<()用来输出类内部的信息：

我们习惯性写法：

template<typename T>class A{public:    A(string value): data(value){}    friend ostream& operator<< (ostream & out, const A<T>& me);private:    T data;}; template<typename T>ostream& operator<< (ostream & out, const A<T>& me){    cout << me.data << endl;    return out;}

GCC编译器警告：

TemplateFriendFunction.cpp:14: warning: friend declaration ‘std::ostream& operator<<(std::ostream&, const A<T>&)’ declares a non-template functionTemplateFriendFunction.cpp:14: note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)

出现问题的原因：定义的友元函数还是一个模版，要实例化为参数变元后才能使用

解决方法有两种：（1）把友元函数体放到类声明内部，使其跟着类的实例化一同实例化：

template<typename T>class A{public:    A(string value): data(value){}    friend ostream& operator<< (ostream & out, const A<T>& me)    {        cout << me.data << endl;        return out;    }private:    T data;};

方法（2）把友元模版放到类模版声明之上，同样在类模版参数化时附带把友元函数参数化：

template<typename T>class A{public:    A(string value): data(value){}    friend ostream& operator<< <>(ostream & out, const A<T>& me);private:    T data;}; template<typename T>ostream& operator<< (ostream & out, const A<T>& me){    cout << me.data << endl;    return out;}

注意在函数名后加上<>

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