如何在检测成语中要求精确的函数签名？

Question

假设我有一个类型T,我想检测它是否有一个下标操作符,我可以用另一种类型调用它Index.以下示例工作得很好:

#include <type_traits>
#include <vector>

template < typename T, typename Index >
using subscript_t = decltype(std::declval<T>()[std::declval<Index>()]);

int main()
{
    using a = subscript_t< std::vector<int>, size_t >;
    using b = subscript_t< std::vector<int>, int    >;
}

但是,当且仅当函数签名完全匹配时,我希望检测函数.在上面的例子中,我希望语句subscript_t< std::vector<int>, int >;抛出一个错误no viable overloaded operator[],因为下标运算符的签名std::vector是

std::vector<T, std::allocator<T>>::operator[](size_type pos);

size_typeGCC 在哪里unsigned long.如何避免隐式转换从int到size_t要发生？

Answer 1

有is_detected,您可以这样做:

template <typename T, typename Ret, typename Index>
using subscript_t = std::integral_constant<Ret (T::*) (Index), & T::operator[]>;


template <typename T, typename Ret, typename Index>
using has_subscript = is_detected<subscript_t, T, Ret, Index>;

static_assert(has_subscript<std::vector<int>, int&, std::size_t>::value, "!");
static_assert(!has_subscript<std::vector<int>, int&, int>::value, "!");

演示

我在SO文档中写了这个:

is_detected

概括type_trait创作:基于SFINAE有实验性的特征detected_or,detected_t,is_detected.

使用模板参数typename Default,template <typename...> Op和typename ... Args:

• is_detected:别名std::true_type或std::false_type依赖于其有效性Op<Args...>
• detected_t:别名Op<Args...>或nonesuch依赖于有效性Op<Args...>.
• detected_or:struct的别名与value_t其是is_detected,和type其为Op<Args...>或Default根据的有效性的Op<Args...>

可以使用std::void_tSFINAE 实现如下:

namespace detail {
    template <class Default, class AlwaysVoid,
              template<class...> class Op, class... Args>
    struct detector
    {
        using value_t = std::false_type;
        using type = Default;
    };

    template <class Default, template<class...> class Op, class... Args>
    struct detector<Default, std::void_t<Op<Args...>>, Op, Args...>
    {
        using value_t = std::true_type;
        using type = Op<Args...>;
    };

} // namespace detail

// special type to indicate detection failure
struct nonesuch {
    nonesuch() = delete;
    ~nonesuch() = delete;
    nonesuch(nonesuch const&) = delete;
    void operator=(nonesuch const&) = delete;
};

template <template<class...> class Op, class... Args>
using is_detected =
    typename detail::detector<nonesuch, void, Op, Args...>::value_t;

template <template<class...> class Op, class... Args>
using detected_t = typename detail::detector<nonesuch, void, Op, Args...>::type;

template <class Default, template<class...> class Op, class... Args>
using detected_or = detail::detector<Default, void, Op, Args...>;

然后可以简单地实现检测方法存在的特征:

template <typename T, typename ...Ts>
using foo_type = decltype(std::declval<T>().foo(std::declval<Ts>()...));

struct C1 {};

struct C2 {
    int foo(char) const;
};

template <typename T>
using has_foo_char = is_detected<foo_type, T, char>;

static_assert(!has_foo_char<C1>::value, "Unexpected");
static_assert(has_foo_char<C2>::value, "Unexpected");

static_assert(std::is_same<int, detected_t<foo_type, C2, char>>::value,
              "Unexpected");

static_assert(std::is_same<void, // Default
                           detected_or<void, foo_type, C1, char>>::value,
              "Unexpected");
static_assert(std::is_same<int, detected_or<void, foo_type, C2, char>>::value,
              "Unexpected");