c++ - 以相反的顺序迭代可变模板参数

Question

如果我手动颠倒传递给它的模板参数的顺序，下面的代码是有效的:

template<typename HeadTag, typename... TailTag>
struct Mapped_scope_deep : public Mapped_scope_deep<TailTag...> {
    typedef typename boost::mpl::at<typename Mapped_scope_deep<TailTag...>::type::type_map,
                                    HeadTag>::type type;
};

template<typename HeadTag>
struct Mapped_scope_deep<HeadTag> {
    typedef typename boost::mpl::at<type_map, HeadTag>::type type;
};

例子:

// typename Mapped_scope_deep<T0, T1, T2, T3>::type
// needs to be written as
typename Mapped_scope_deep<T3, T2, T1, T0>::type

我已尝试在此处解决此问题:

template<typename map, typename HeadTag, typename... TailTag>
struct Mapped_scope_deep_r :
    public Mapped_scope_deep_r< typename boost::mpl::at<map, HeadTag>::type::type_map, TailTag...> {
  typename Mapped_scope_deep_r< typename boost::mpl::at<map, HeadTag>::type::type_map, TailTag...>::type type;
};

template<typename map, typename HeadTag>
struct Mapped_scope_deep_r<map, HeadTag> {
    typedef typename boost::mpl::at<map, HeadTag>::type type;
};

template<typename... Tags>
struct Mapped_scope_deep3 :
    public Mapped_scope_deep_r<type_map, Tags...> {
    typedef typename Mapped_scope_deep_r<type_map, Tags...>::type type;
};

例子:

typename Mapped_scope_deep<T0, T1, T2, T3>::type

但这以编译错误结束:

./compressed_enums.hxx:197:66: error: typename specifier refers to non-type member 'type' in 'Gamblify::Asdf<unsigned char, CAT>::Mapped_scope_deep_r<boost::mpl::map<boost::mpl::pair<Cat, Gamblify::Category2<unsigned char, 1, Cat, B_First, TagA_array_2, B_Second> > >, Cat, First>'
    typedef typename Mapped_scope_deep_r<type_map, Tags...>::type type;
            ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~

我做错了什么，它们是一种更简单的以相反顺序进行类似折叠操作的方法吗？

Answer 1

您在 Mapped_scope_deep_r 中缺少一个 typedef。这一行声明了一个对象，而不是一个类型:

typename Mapped_scope_deep_r< typename boost::mpl::at<map, HeadTag>::type::type_map, TailTag...>::type type;

至于反转包的顺序，有一些肮脏的技巧，但最好的方法是定义一个元函数 tuple_reverse 并用它来过滤模板的输入。