Question

我正在尝试重载模板和继承类中的ostream运算符，我一直在关注一些提示here和here，但是我得到了重新定义错误。这是我的代码的再现：

#include <iostream>

enum type
{
        A,
        B
};

template <type T>
class base
{
protected:
        virtual std::ostream& print(std::ostream& out) const =0;
};

template <type T>
class derived: public base<T>
{
protected:
        virtual std::ostream& print(std::ostream& out) const
        {
                out<<"Hello World.\n";
                return out;
        }
public:
        template <type S>
        friend std::ostream& operator<<(std::ostream& out, const derived<S>& D)
        {
                return (D.print(out));
        }
};

int main ()
{
#ifdef __NOT_WORKING__
        derived<A> a;
        std::cout<<a;
        derived<B> b;
        std::cout<<b;
#else
        derived<A> a;
        std::cout<<a;
#endif
        return 0;
}

如果我只定义派生的A类，一切正常，但如果我定义派生的A和派生的B类，我会从编译器中得到这个错误：

test.cpp: In instantiation of 'class derived<(type)1u>':
test.cpp:38:20:   required from here
test.cpp:27:30: error: redefinition of 'template<type S> std::ostream& operator<<(std::ostream&, const derived<S>&)'
         friend std::ostream& operator<<(std::ostream& out, const derived<S>& D)
                              ^
test.cpp:27:30: note: 'template<type S> std::ostream& operator<<(std::ostream&, const derived<S>&)' previously defined here
test.cpp: In instantiation of 'std::ostream& operator<<(std::ostream&, const derived<S>&) [with type S = (type)1u; type T = (type)0u; std::ostream = std::basic_ostream<char>]':
test.cpp:39:20:   required from here
test.cpp:20:31: error: 'std::ostream& derived<T>::print(std::ostream&) const [with type T = (type)1u; std::ostream = std::basic_ostream<char>]' is protected
         virtual std::ostream& print(std::ostream& out) const
                               ^
test.cpp:29:37: error: within this context
                 return (D.print(out));
                                     ^

为什么要重新定义好友功能？谢谢你的时间。

PS。我正在使用gcc49。

Answer 1

替换

template <type S>
friend std::ostream& operator<<(std::ostream& out, const derived<S>& D)
{
    return (D.print(out));
}

带

friend std::ostream& operator<<(std::ostream& out, const derived<T>& D)
{
    return (D.print(out));
}

错误将消失。

使用早期定义，您尝试定义具有相同签名的新模板函数。

Answer 2

[temp.friend] / 4：

在类中的友元函数声明中定义函数时模板，当函数使用odr时，该函数被实例化。 对多个声明和定义的相同限制适用于非模板函数的声明和定义也适用这些隐含的定义。

Clang汇编上述罚款 - 这可以被视为一个错误，但回想一下，违反ODR的行为是错误的，不需要诊断。

您有两种方法可以解决这个问题。提取模板的定义：

template <typename T>
class derived: public base<T>
{
//     [..]

       template <type S>
       friend std::ostream& operator<<(std::ostream& out, const derived<S>& D);
};

template <type S>
std::ostream& operator<<(std::ostream& out, const derived<S>& D)
{
    return (D.print(out));
}

或者让操作员成为非模板：

template <type T>
class derived: public base<T>
{
//     [..]

       friend std::ostream& operator<<(std::ostream& out, const derived<T>& D)
       {
               return (D.print(out));
       }
};

模板和继承类中的ostream重载中的重新定义错误

2 个答案: