c++ - 专门化成员 S::display 需要 ‘template<>’ 语法

Question

我正在创建一个特征类来帮助我的程序。我有一个名为 operations 的模板类包含方法 display和 area .当我定义这些函数时，我得到了错误。他们在这里:

error: specializing member ‘traits::operations<Rectangle>::display’ requires ‘template<>’ syntax
error: specializing member ‘traits::operations<Rectangle>::area’ requires ‘template<>’ syntax

如您所见，编译器要我插入 template <>就在这些定义之前。但是当我这样做时，我会收到一大页错误。出了什么问题，我该如何解决？

这是我的程序。

namespace traits
{
    template <typename P>
    struct operations
    {
        static void display(Rectangle const &, std::ostream &);
        static void area(Rectangle const &);
    };

    template <typename P, int N>
    struct access {};
}

namespace traits
{
    template <int N>
    struct access<Rectangle, N>
    {
        static double get(Rectangle const &);
    };
}

// The errors occur here
namespace traits
{
    static void operations<Rectangle>::display(Rectangle const &rect, std::ostream &os)
    {
        os << rect.width  << '\n';
        os << rect.height << '\n';
        os << area(rect)  << '\n'; 
    }

    static void operations<Rectangle>::area(Rectangle const& rect)
    {
        double width =  get<0>(rect);
        double height = get<1>(rect);

        return width * height;
    }
}

namespace traits
{
    template <>
    struct access<Rectangle, 0>
    {
        static double get(Rectangle const &rect)
        {
            return rect.width;
        }
    };

    template <>
    struct access<Rectangle, 1>
    {
        static double get(Rectangle const &rect)
        {
            return rect.height;
        }
    };
}

template <int N, typename P>
static inline double get(P const &p)
{
    return traits::access<P, N>::get(p);
}

template <typename P>
static inline void display(P const &p)
{
    traits::operations<P>::display(p, std::cout);
}

template <typename P>
static inline double area(P const &p)
{
    return traits::operations<P>::area(p);
}

int main()
{

}

这是一个显示错误的程序 - http://ideone.com/WFlnb2#view_edit_box

感谢任何帮助。

---

感谢评论的帮助，我摆脱了这两个错误，但在添加 template<> 后我并没有得到更多声明并修复 area 的返回类型:

error: cannot declare member function ‘static void traits::operations<P>::display(const Rectangle&, std::ostream&) [with P = Rectangle; std::ostream = std::basic_ostream<char>]’ to have static linkage [-fpermissive]
error: explicit template specialization cannot have a storage class
error: specialization of ‘static double traits::operations<P>::area(const Rectangle&) [with P = Rectangle]’ after instantiation
error: explicit template specialization cannot have a storage class

Answer 1

您的功能:display和 area应该这样写:

    template <>
    double operations<Rectangle>::area( Rectangle const& rect )
    {
            double width =  get<0>(rect);
            double height = get<1>(rect);

            return width * height;
    }

• 至于模板专用函数，template <>应该放在在函数的头部。
• 对于静态成员函数，static不应该出现在函数的定义体。