为什么 std::void_t 在这种情况下不起作用

Question

#include <type_traits>
template<typename, typename = void>
struct IsIterator final : std::false_type
{};
template<typename T>
struct IsIterator<T,
    std::void_t<std::enable_if_t<std::is_base_of_v<std::input_iterator_tag,
        typename std::iterator_traits<T>::iterator_category>>>>
    final : std::true_type
{};

int main()
{
    return IsIterator<void*>::value;
}

Clang 8.0 给出以下错误消息：

/usr/bin/../include/c++/v1/iterator:507:16: error: cannot form a reference to 'void'
    typedef _Tp& reference;
               ^
main.cpp:20:23: note: in instantiation of template class 'std::__1::iterator_traits<void *>' requested
      here
        typename std::iterator_traits<T>::iterator_category>>>>
                      ^
main.cpp:29:16: note: during template argument deduction for class template partial specialization
      'IsIterator<T, std::void_t<std::enable_if_t<std::is_base_of_v<std::input_iterator_tag, typename
      std::iterator_traits<T>::iterator_category> > > >' [with T = void *]
        return IsIterator<void*>::value;
               ^
main.cpp:29:16: note: in instantiation of template class 'IsIterator<void *, void>' requested here

为什么在这种情况下std::void_t不起作用？

Answer 1

std::iterator_traits<T>::iterator_category强制实例化格式错误的std::iterator_traits<T>(SFINAE 的硬错误而不是软错误(void*。

您必须手动处理void*。