Question

我有一些深度C ++ 11可变参数模板的魔术用法代码。我使用专门的类来识别模板树的部分，我命名为“标签”。每个标签都有唯一的编号标识符我需要找到一种方法来访问标签的标识符

以下是一些小例子：

#include <string>
#include <iostream>

template <unsigned long ID, typename PARAM>
class tag_
{
    //getter for ID. Better to exclude it. If possible - move to external helper
    constexpr static unsigned long get_id() {return ID;}
    //PARAM is used in some functions of this class
};

//tag's declarations for future usage
template<typename PARAM>
using tag1 = tag_<1UL, PARAM>;
template<typename PARAM>
using tag2 = tag_<2UL, PARAM>;
template<typename PARAM>
using tag3 = tag_<3UL, PARAM>;

//helper class that can iterate TAGS
template <template<typename> class... TAGS>
struct helper
{};
template <template<typename> class TAG>
struct helper<TAG>
{
    static void print_tag(std::ostream& out)
    {
        out << std::string("Tag");
        out << TAG::get_id();  // Here I can't call: error: 'template<class> class TAG' used without template parameters
    }
};
template <template<typename> class TAG, template<typename> class... TAGS>
struct helper<TAG, TAGS...>
{
    static void print_tag(std::ostream& out)
    {
        out << std::string("Tag");
        out << TAG::get_id();  // Here I can't call: error: 'template<class> class TAG' used without template parameters
        out << std::string(", ");
        helper<TAGS...>::print_tag(out);
    }
};

//this class uses tags for some processing
template <template<typename> class... TAG_LIST>
class tagged
{
public:
    void test1(std::ostream& out)
    {
        out << std::string("This function works good");
    }
    void test2(std::ostream& out)
    {
        helper<TAG_LIST...>::print_tag(out);
    }
};

// this class is re-defined for some types of T
template<typename T, typename PARAM>
class usage
{
public:
    void test1(std::ostream& out)
    {
        details.test1(out);
    }
    void test2(std::ostream& out)
    {
        details.test2(out);
    }

    T details;
    PARAM params;
};

//endpoint
struct finish{};

//definition for future usage
template<template<typename> class T1, template<typename> class T2, template<typename> class T3, typename PARAM>
using multitag = usage<tagged<T1, T2, T3>, PARAM>;

int main(int argc, const char* argv[])
{
    // this way I am construction my objects tree
    multitag<tag1, tag2, tag3, tag1<tag2<tag3<finish>>>> tmp;
    tmp.test1(std::cout); // ok
    std::cout << std::endl;
    tmp.test2(std::cout);  //compile error. I want to get "Tag1, Tag2, Tag3" printed
    std::cout << std::endl;
}

Answer 1

multitag<tag1, tag2, tag3, tag1<tag2<tag3<finish>>>> tmp;

这是您对tmp的定义，您只需传入tag1，这是一个模板类型别名

template<typename PARAM>
using tag1 = tag_<1UL, PARAM>;

但是未提供其模板参数PARAM。所以当你想要使用它时，

out << TAG::get_id()

你应该给它一个参数，例如

out << TAG<dummy>::get_id();

dummy可以是任何内容，例如

class dummy{};

http://coliru.stacked-crooked.com/a/bb1d924dc3d5b774

Answer 2

最接近的是为每个tagX声明一个专门化：

template <template <typename> class TAG> struct tag_id;

template <> struct tag_id<tag1> : std::integral_constant<unsigned long, 1UL> {};
template <> struct tag_id<tag2> : std::integral_constant<unsigned long, 2UL> {};
template <> struct tag_id<tag3> : std::integral_constant<unsigned long, 3UL> {};

out << tag_id<TAG>{};

Set Up the NativeScript CLI on OS X

使用部分类型知识访问模板的参数

2 个答案: