模板和继承类中的 ostream 重载重定义错误答案

【问题标题】：Redefinition error in ostream overload in template and inherited classes模板和继承类中的 ostream 重载重定义错误
【发布时间】：2015-02-27 11:21:11
【问题描述】：

我正在尝试在模板和继承类中重载 ostream 运算符，并且我一直在遵循一些提示 here 和 here，但我收到了重新定义错误。这是我的代码的复制：

#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;
}

如果我只定义一个派生 A 类，一切正常，但如果我定义一个派生 A 和一个派生 B 类，我会从编译器收到此错误：

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？感谢您的宝贵时间。

PS。我正在使用 gcc49。

【问题讨论】：

标签： c++ templates inheritance ostream

【解决方案1】：

替换

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

与

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

错误会消失。

使用较早的定义，您正在尝试定义具有相同签名的新模板函数。

【讨论】：

【解决方案2】：

[temp.friend]/4:

当函数在类的友元函数声明中定义时模板，当函数被使用时，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声明和定义也适用这些隐含的定义。

Clang 编译了上述罚款 - 这可以被认为是一个错误，但请记住，违反 ODR 是不正确的，不需要诊断。

你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));
}

或者将操作符设为非模板：

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

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

【讨论】：