VS2017上SFINAE的編譯錯誤

Question

#include <type_traits>
#define str_cat(first, second) first##second
#define has_xxx(member_name) \
template<class T, class U = void> \
struct str_cat(has_, member_name): std::false_type {};\
template<class T>\
struct str_cat(has_, member_name)<T, typename SFINAE<typename T::member_name>::type>\
    : std::true_type {};\
template<class T>\
struct str_cat(has_, member_name)<T, typename SFINAE<decltype(T::member_name)>::type>\
    : std::true_type {};

template<class T>
struct SFINAE {
    using type = void;
};

struct A {
    int i = 0;
    void foo() {}
    using void_t = void;
};

struct B {
    int j = 0;
    void goo() {}
    using void_t = void;
};

has_xxx(i)
has_xxx(foo)
has_xxx(j)
has_xxx(goo)
has_xxx(void_t) //compile error if `has_xxx(i)` appears at the head

int main()
{

    //has_i<A>::value; // true
    //has_i<B>::value; // false

    has_foo<A>::value; // true
    has_foo<B>::value; // false
    has_goo<B>::value; // true

    has_void_t<A>::value; // true
    has_void_t<B>::value; // true

    return 0;
}

在VS2017上無法編譯
https://gcc.godbolt.org/z/JkOhLi
錯誤：C2752'模板'：多個部分專業化與模板參數列表匹配。
但是在gcc和clang上還可以。 http://coliru.stacked-crooked.com/a/6b9490f6b127ae88
如果更改宏的順序，它將編譯：

has_xxx(foo)
has_xxx(i) //now compiles successfully
has_xxx(j)
has_xxx(goo)
has_xxx(void_t) //compile error if `has_xxx(i)` appears at the head

或者只是更改結構A中的成員名稱：

struct A {
    int k = 0; // i -> k, now compiles successfully !!!!
    void foo() {}
    using void_t = void;
};

我不知道原因。
宏的順序是否重要還是SFINAE上的MSVC錯誤？

Answer 1

它看起來像是MSVC錯誤，但是可以像這樣輕松避免：

#include <type_traits>

#define str_cat(first, second) first##second
#define custom_trait(trait_name, expr) \
  template<class T, class U = void> \
  struct trait_name: std::false_type {}; \
  template<class T> \
  struct trait_name<T, std::void_t<expr>> : std::true_type {};
#define has_xxx(member_name) \
  custom_trait(str_cat(has_type_, member_name), typename T::member_name) \
  custom_trait(str_cat(has_value_, member_name), decltype(T::member_name)) \
  template<class T>\
  using str_cat(has_, member_name) = \
      std::bool_constant<str_cat(has_type_, member_name)<T>::value \
                      || str_cat(has_value_, member_name)<T>::value>;

注意 ：此代碼（以及您的代碼）不允許您檢測方法。

我建議您報告問題（ Help -> Send Feedback -> Report a problem是Visual Studio）。