c++ - 模板类友元运算符成员函数

Question

我试图让一个模板化类中的友元函数进行编译，但我不明白错误消息和警告。我已经对这个问题进行了演示。我得到的错误是:

prog.cpp:8:57: error: non-class, non-variable partial specialization C operator+(const B& lhs, const C& rhs);

prog.cpp:15:59: warning: friend declaration 'C operator+(const B&, const C&)' declares a non-template function [-Wnon-template-friend] friend C operator+(const B& lhs, const C& rhs);

prog.cpp:15:59: note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)

#include <iostream>
using namespace std;

template<typename A, typename B>
class C;

template<typename A, typename B>
C<A, B> operator+<A, B>(const B& lhs, const C<A, B>& rhs);

template<typename A, typename B>
struct C
{
    A val_;
    C operator+(const C& other) const;
    friend C<A, B> operator+(const B& lhs, const C<A, B>& rhs);
};

template<typename A, typename B>
C<A, B> C<A, B>::operator+(const C<A, B>& other) const
{
    C<A, B> c;
    c.val_ = this->val_ + other.val_;
    return c;
}

template<typename A, typename B>
 C<A, B> operator+(const B& lhs, const C<A, B>& rhs)
{
    C<A, B> c;
    c.val_ = lhs + rhs.val_;
    return c;
}

int main() 
{
    C<string, char> c0,c1;
    c0.val_ = " C0 ";
    c1.val_ = " C1 ";
    cout << "Stuct:" << (c0 + c1).val_ << '\n';
    cout << "Friend:" << ('~' + c1).val_ << endl;
    return 0;
}

Answer 1

最简单的是在类中内联代码:

template <typename A, typename B>
struct C
{
    A val_;
    C operator+(const C& other) const
    {
        C c;
        c.val_ = this->val_ + other.val_;
        return c;
    }

    friend C operator+ (const B& lhs, const C& rhs)
    {
        C c;
        c.val_ = lhs + rhs.val_;
        return c;
    }
};

Demo

类中没有内联的代码，需要特别注意声明的前向声明顺序，奇怪语法<> :

template <typename A, typename B> struct C;

template <typename A, typename B>
C<A, B> operator+ (const B& lhs, const C<A, B>& rhs);

template <typename A, typename B>
struct C
{
    A val_;

    friend C<A, B> operator+<> (const B& lhs, const C<A, B>& rhs);

    C operator+(const C& other) const;
};


template <typename A, typename B>
C<A, B> operator+ (const B& lhs, const C<A, B>& rhs)
{
    C<A, B> c;
    c.val_ = lhs + rhs.val_;
    return c;
}

template <typename A, typename B>
C<A, B> C::operator+(const C<A, B>& other) const
{
    C<A, B> c;
    c.val_ = this->val_ + other.val_;
    return c;
}

Demo