专业化成员S :: display需要'template<&''语法 [英] specializing member S::display requires ‘template<>’ syntax

问题描述

我正在创建一个trait类来帮助我的程序。我有一个包含方法 display 和区域的模板类 operations $ c>。当我定义这些功能，我得到错误。这里他们是：

I'm creating a trait class to help with my program. I have a template class called operations that contains the methods display and area. When I define these functions I get errors. Here they are:

错误：specializing member 'traits :: operations< Rectangle& code>需要'template<>'语法

错误：specializing member 'traits :: operations< Rectangle> ; :: area'需要'template<>'语法

如你所见，编译器要求我在这些定义之前插入 template<> 。但是当我做，我得到一个巨大的页面的错误。

As you can see, the compiler wants me to insert template <> just before those definitions. But when I do, I get a huge page of errors. What's going wrong and how can I fix it?

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()
{

}

b
$ b

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

感谢任何一切帮助。

由于评论的帮助我摆脱了这两个错误，但我没有得到更多后添加模板<> 声明和修复的返回类型：

Thanks to help from the comments I got rid of those two errors, but I'm not getting more after adding the template<> declaration and fixing the return type of area:

错误：无法声明成员函数'static void traits :: operation< P> :: display（const Rectangle& std :: ostream&）[with P = Rectangle; std :: ostream = std :: basic_ostream< char>]'具有静态链接[-fpermissive]

错误：显式模板专门化不能有存储类

错误：实例化之后'static double traits :: operations< P> :: area（const Rectangle&）[with P = Rectangle]'
错误：显式模板专门化不能有存储类

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

推荐答案

显示和区域应该这样写：

    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<> 应放置在函数头部的
  。


• 对于静态成员函数， static 不应出现在函数的
  定义体。


• As for template specialized functions, template <> should be placed at the head of the function.
• For static member functions, static should not appear at the definition body of the function.

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