Question

享受boost :: hana的乐趣。我希望检查一个特定的嵌套类型，它的作用类似于另一种类型的标签，所以我借用了hana :: when_valid示例并定义了一个类is_S及其支持SFINAE的特化：

#include <iostream>

#include <boost/hana/core/when.hpp>
namespace hana = boost::hana;

#define V(x) std::cout << x << std::endl

struct S_tag { };

struct S {
    using tag = S_tag;
};

struct T {
    using tag = int;
};

template< typename T, typename = hana::when< true > >
struct is_S {
    static constexpr bool value = false;
};

template< typename T >
struct is_S< T, hana::when_valid< typename T::tag > > {
    static constexpr bool value = std::is_same<
        typename T::tag, S_tag >::value;
};

int main () {
    std::cout << "is_S (    S { }) = "; V ((is_S< S >::value));
    std::cout << "is_S (    T { }) = "; V ((is_S< T >::value));
    std::cout << "is_S (float { }) = "; V ((is_S< float >::value));

    return 0;
}

打印：

$ clang++ -std=c++1z sfinae.cpp && ./a.out | c++filt
is_S (    S { }) = 1
is_S (    T { }) = 0
is_S (float { }) = 0

是否有更简单/更短/更简洁的方式来编写相同的支票，与hana哲学的价值型计算保持一致？

Answer 1

以下是我的写作：

#include <boost/hana.hpp>
#include <iostream>
namespace hana = boost::hana;


struct S_tag { };
struct S { using tag = S_tag; };
struct T { using tag = int; };

auto tag_of = [](auto t) -> hana::type<typename decltype(t)::type::tag> {
    return {};
};

auto is_S = [](auto t) {
    return hana::sfinae(tag_of)(t) == hana::just(hana::type<S_tag>{});
};

int main() {
    std::cout << "is_S (    S { }) = " << is_S(hana::type<S>{})() << std::endl;
    std::cout << "is_S (    T { }) = " << is_S(hana::type<T>{})() << std::endl;
    std::cout << "is_S (float { }) = " << is_S(hana::type<float>{})() << std::endl;
}

Answer 2

我受到诱惑：

template<class...T>
constexpr std::integral_constant<bool,false> is_S(T const&...){ return {}; }
template<class T>
constexpr
std::integral_constant<bool,std::is_same<typename T::tag,S_tag>{}>
is_S(T const&){ return {}; }

使用boost hana检查特定的嵌套类型/标记

2 个答案: