在类型匹配中忽略模板参数

Question

这是我当前的程序：

#include <type_traits>

template<class Feature, class... FeatureList>
struct has_feature {
    static constexpr bool value = (std::is_same_v<Feature, FeatureList> || ...);
};

template<class Feature, class... FeatureList>
inline constexpr bool has_feature_v = has_feature<Feature, FeatureList...>::value;

template<class Feature, class ...FeatureList>
static constexpr bool isConfiguredWith() {
    return has_feature_v<Feature, FeatureList...>;
}

struct CanWalk {
};

struct CanNotWalk {
};

template<class... FeatureList>
struct Robot {
    static auto configure() {
        return Robot<WalkFeature < FeatureList...>>
        ();
    }

private:
    template<typename ...Config>
    using WalkFeature =
    std::conditional_t<isConfiguredWith<CanWalk, Config...>(), CanWalk, CanNotWalk>;
};

int main() {
    Robot<CanWalk> robot_A = Robot<CanWalk>::configure();
    Robot<CanNotWalk> robot_B = Robot<>::configure();

    return 0;
}

基本上，Robot是一个struct，可以配置很多其他的struct（这里用它们作为token），然后Robot<T...>::configure() trim和组织传入的模板参数至 Robot。最后我们有：

    Robot<CanWalk> robot_A = Robot<CanWalk, CanWalk>::configure();
    Robot<CanNotWalk> robot_B = Robot<>::configure();

虽然重复的特征CanWalk作为模板参数传入，但在通过函数configure()构造Robot时，它们都被删除了。

在我将模板参数添加到功能 CanWalk:

之前，它运行良好

template <int Speed>
struct CanWalk {
};

现在一切都崩溃了，因为 CanWalk 不再是合法类型，它需要一个模板参数。

错误 error: use of class template 'CanWalk' requires template arguments 发生自：

template<typename ...Config>
using WalkFeature =
std::conditional_t<isConfiguredWith<CanWalk>(), CanWalk, CanNotWalk>;

我该如何解决？

如何将它们定义为：

Robot<CanWalk<5>> robot_A = Robot<CanWalk<5>>::configure();
Robot<CanNotWalk> robot_B = Robot<>::configure();

?

实时代码：https://godbolt.org/z/4K9TxohWq

Answer 1

我在这里看到三个挑战：

您需要从参数包中提取特定特征 (CanWalk)，而不考虑其参数
您需要忽略相同特征的额外副本
如果该特征不存在，您需要设置默认特征 (CanNotWalk)

我不知道有什么比递归更好的方法了：

// definition
template<template<auto...> class FeatureType, class DefaultFeature, class... FeatureList>
struct find_feature;

// next trait matches, stop recursing and return it
template<template<auto...> class FeatureType, class DefaultFeature, auto... Param, class... RemainingFeatures>
struct find_feature<FeatureType, DefaultFeature, FeatureType<Param...>, RemainingFeatures...> {
    using type = FeatureType<Param...>;
};

// next trait does not match, skip by inheriting from rest of list
template<template<auto...> class FeatureType, class DefaultFeature, class FirstFeature, class... RemainingFeatures>
struct find_feature<FeatureType, DefaultFeature, FirstFeature, RemainingFeatures...> 
    : find_feature<FeatureType, DefaultFeature, RemainingFeatures...> { };

// no more traits, return default trait
template<template<auto...> class FeatureType, class DefaultFeature>
struct find_feature<FeatureType, DefaultFeature> {
    using type = DefaultFeature;
};

// alias
template<template<auto...> class FeatureType, class... FeatureList>
using find_feature_t = typename find_feature<FeatureType, FeatureList...>::type;

这应该适用于 class 模板和任意数量的非类型参数的任何功能。

用法：

template <int speed>
struct CanWalk {};
struct CanNotWalk {};

template<class... FeatureList>
struct Robot {
    static auto configure() {
        return Robot<WalkFeature < FeatureList...>>
        ();
    }

private:
    template<typename ...Config>
    using WalkFeature = find_feature_t<CanWalk, CanNotWalk, Config...>;
};

auto robot_A = Robot<CanWalk<42>, CanWalk<25>>::configure();
static_assert(std::is_same_v<decltype(robot_A), Robot<CanWalk<42>>>, "");

auto robot_B = Robot<>::configure();
static_assert(std::is_same_v<decltype(robot_B), Robot<CanNotWalk>>, "");

Answer 2

删除一些通用性，您可以这样做：

template <typename T> struct Tag { using type = T; };

template <std::size_t> struct CanWalk {};
struct CanNotWalk {};

template <std::size_t N> Tag<CanWalk<N>> has_walk(Tag<CanWalk<N>>); // No impl
template <typename T> Tag<CanNotWalk> has_walk(Tag<T>); // No impl

template<class... FeatureList>
struct Robot {
    static auto configure() {
        struct all_features : Tag<FeatureList>..., Tag<struct Empty> {};
        return Robot<typename decltype(has_walk(all_features{}))::type>{};
    }
};

int main() {
    [[maybe_unused]] Robot<CanWalk<42>> robot_A = Robot<CanWalk<42>>::configure();
    [[maybe_unused]] Robot<CanNotWalk> robot_B = Robot<>::configure();
}

Demo

这种简单的方法不能处理重复项，但可以做一些额外的工作（想法是让 struct all_features : Tag<Ts, Is>... 和 std::index_sequence）

在类型匹配中忽略模板参数

ignore template parameter in type matching

c++

templates

template-meta-programming