获取模板列表的第一个模板的元函数 [英] Meta-Function to get the first template of a list of templates

问题描述

使用以下元函数 front 获取类型列表的第一个类型，我尝试编写一个类似的元函数来提取模板列表的第一个模板.

Using the following metafunction front to get the first type of a typelist I try to write a similar metafunction to extract the first template of a list of templates.

namescpace detail {
    template<typename L> struct front_impl;
    template<template<typename, typename...> typename L, typename F, typename... I>
    struct front_impl<L<F, I...>> {
        typedef F type;  
    };
}
template<typename L> struct front_impl;
template<template<typename, typename...> typename L, typename F, typename... I>
struct front_impl<L<F, I...>> {
    typedef F type;  
};

您可以按如下方式使用它:

One can use this as follows:

template<typename... T>
struct List {
    inline static constexpr size_t size = sizeof...(T);
};

using l1 = List<A, B, C>;
using f1 = front<l1>;

现在我尝试对模板列表做同样的事情.因此我使用了一个模板列表 TList:

Now I try to do the same with a list of templates. Therefore I use a list of templates TList:

template<template<typename> typename... TT>
struct TList {
    inline static constexpr size_t size = sizeof...(TT);
};

和元函数tfront:

template<typename T> struct tfront;
template<template<typename> typename F, template<typename> typename... R>
struct tfront<TList<F, R...>> {
    template<typename T> using type = F<T>;
};

然后我可以提取模板列表中的第一个 A、B、...(此处未显示):

Then I can extract the first of a list of templates A, B, ... (not shown here):

    using tlist = TList<A, B, C>;

    template<typename T>
    using f = typename tfront<tlist>::template type<T>;

    f<int> xx;

然后 xx 属于 A 类型.

Then xx ist of type A<int>.

问题是:我可以用与 front 相同的方式编写元函数 tfront，这不是对模板列表 TList 但是 模板的每个可变参数模板?所以我想为 tfront 引入一个参数 L 作为带有可变模板列表的模板模板参数，这样编译器也必须推导出输入 L 就像 front 一样.

The Question is: can I write the metafunction tfront in the same way as front, that is not as a partial specialization for the list of templates TList but for every variadic template of templates? So I would like to introduce a parameter L for tfront as a template-template-parameter with a variadic list of templates, so that the compiler must also deduce the type of the type L as in the case of front.

我想写一些类似的东西(但用作什么???):

I would like to write something like (but what to use as ???):

template<template<typename> typename F, ??? TL, template<typename> typename... R>
struct tfront<TL<F, R...>> {
    template<typename T> using type = F<T>;
};

推荐答案

你需要额外一层 template类型名称...:

template<typename>
struct tfront;

template<template<typename> typename F, 
         template<template<typename> typename...> typename TL, 
         template<typename> typename... R>
struct tfront<TL<F, R...>> 
{
};

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