MSVC无法编译SFINAE检查

Question

我应该注意一些编译器标志吗？下面的代码可以正常工作，并且可以按照GCC和Clang的要求运行，但不适用于MSVC。是什么使std::allocator的分配在这里如此不同？

#include <type_traits>
#include <memory>

template <typename T, typename = std::void_t<>>
constexpr bool declares_allocate = false;

template <typename T>
constexpr bool declares_allocate<T, std::void_t<decltype(&T::allocate)>> = true;

struct normal_struct {

    auto allocate() -> void {}

};

template <typename T>
struct struct_template {

    auto allocate() -> void {}

};

template <typename T>
class class_template_public {

    public:
        auto allocate() -> void {}

};

template <typename T>
class class_template_private {

    private:
        auto allocate() -> void {}

};


auto main() -> int {

    auto allocator = std::allocator<float>();
    auto memory = allocator.allocate(1024);

    static_assert(declares_allocate<normal_struct>);                    // pass
    static_assert(declares_allocate<struct_template<float>>);           // pass
    static_assert(declares_allocate<class_template_public<float>>);     // pass
//  static_assert(declares_allocate<class_template_private<float>>);    // fails
    static_assert(declares_allocate<std::allocator<float>>);            // fails when compiled by MSVC but not Clang or GCC

    allocator.deallocate(memory, 1024);

    return 0;

}

https://godbolt.org/z/GVyNQZ

Answer 1

取决于C ++标准实现，可以将std::allocator<T>::allocate定义为：

---

T* allocate(size_type n, const void* hint = 0);

---

T* allocate(size_type n);
T* allocate(size_type n, const void* hint);

---

T* allocate(size_t n);

---

即，第二种实现使allocate成为重载的成员函数。现在MSVC's standard library中使用了那个：

_NODISCARD __declspec(allocator) _Ty* allocate(_CRT_GUARDOVERFLOW const size_t _Count) {
    return static_cast<_Ty*>(_Allocate<_New_alignof<_Ty>>(_Get_size_of_n<sizeof(_Ty)>(_Count)));
}

_CXX17_DEPRECATE_OLD_ALLOCATOR_MEMBERS _NODISCARD __declspec(allocator) _Ty* allocate(
    _CRT_GUARDOVERFLOW const size_t _Count, const void*) {
    return allocate(_Count);
}

这种实现使&T::allocate模棱两可，因此在替换期间被拒绝。