c – 如何检测成员函数是否具有特定功能,采取特定参数?

我已经有SFINAE代码来检测给定的成员函数是否存在并且可以接受特定类型的实例.

我试图检测何时有一个特定的成员函数,因为它的参数是非参考.

示例代码

template<typename TYPE_T>
class HasSwapMemberImpl
{
    /*
     * This uses the compilers type deduction magic to ensure that there's SOME swap function
     * that's a member of CLASS_T, that can be passed an instance of CLASS_T, in some fashion.
     * This doesn't determine that the argument for the function is a reference.
     */
    template <typename CLASS_T>
    static auto test_compatible_swap_function_exists(CLASS_T * p) -> decltype(p->swap(*static_cast<CLASS_T*>(nullptr)), boost::true_type());

    /*
     * If no substitutions can satisfy p->swap(*static_cast<CLASS_T*>(nullptr)
     * we end up here as fallback.
     */
    template<typename>
    static boost::false_type test_compatible_swap_function_exists(...);

public:
    typedef decltype(test_compatible_swap_function_exists<TYPE_T>(nullptr)) type;
    static const bool value = type::value;
};

/**
 * \brief This MetaProgramming helper class determines if there exists a method named "swap" in the ARG_T type, that takes ARG_T as an argument.
 *
 * If ARG_T.swap(ARG_T) is a valid function, then HasSwapMember inherits from boost::true_type. Else it inherits from boost::false_type.
 */
template<typename ARG_T>
struct HasSwapMember : public HasSwapMemberImpl<ARG_T>::type { };

这适用于我写过的几十个测试用例,所以我在这里不需要任何帮助.

相反,我正在尝试做的是检测这样的情况

struct NonReferenceSwap
{
    void swap(NonReferenceSwap) {}
};

struct ReferenceSwap
{
    void swap(ReferenceSwap &) {}
};

我想要一些元编程助手MetaHelperType,这样MetaHelperType继承自boost :: true_type,但是MetaHelperType会导致编译器错误地输出“交换函数必须带参考参数!”这样的消息.

或者,换句话说:

MetaHelperType<ReferenceSwap>; // IS A boost::true_type
MetaHelperType<NoSwapFunction>; // IS A boost::false_type
MetaHelperType<NonReferenceSwap>; // Compiler error

我为什么要这个?当bar作为值传递时,尝试调用foo.swap(bar)是一个无意义的概念.我的组织有很多新手c程序员,我们希望能够立即发现这个错误,而不是发现有人忘记了&两个月后.

我正在使用Visual Studio 2010,因此并非所有c 11功能都可用.上面的代码确实可以正常工作,所以至少这些功能按预期工作.

这可能吗?如果是这样,需要什么最低版本的C? C 11,C 14,C 17?

最佳答案 我使用以下内容进行精确签名检查:

#include <cstdint>

#define DEFINE_HAS_SIGNATURE(traitsName, funcName, signature)               \
    template <typename U>                                                   \
    class traitsName                                                        \
    {                                                                       \
    private:                                                                \
        template<typename T, T> struct helper;                              \
        template<typename T>                                                \
        static std::uint8_t check(helper<signature, &funcName>*);           \
        template<typename T> static std::uint16_t check(...);               \
    public:                                                                 \
        static                                                              \
        constexpr bool value = sizeof(check<U>(0)) == sizeof(std::uint8_t); \
    }

DEFINE_HAS_SIGNATURE(has_ref_swap, T::swap, void (T::*)(T&));
DEFINE_HAS_SIGNATURE(has_value_swap, T::swap, void (T::*)(T));

然后,您可以编写助手检查:

template <typename T>
struct MetaHelperType_impl
{
    static_assert(!has_value_swap<T>::value,
                  "Incorrect implementation of T::swap, "
                  "signature should be void T::swap(T&) instead of T::swap(T)");

    using type = std::conditional_t<has_ref_swap<T>::value,
                                    boost::true_type,
                                    boost::false_type>;
};

template <typename T>
using MetaHelperType = typename MetaHelperType_impl<T>::type;

Demo

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