c++ - clang : candidate template ignored: substitution failure: typedef 'type' cannot be referenced with a class specifier

Question

与 GCC 5 相比，Clang 6 提示以下错误:

candidate template ignored: substitution failure [with U = char, Us = ]: typedef 'type' cannot be referenced with a class specifier Tuple(U&& u, Us&&... rest) : m_element(::std::forward(u)), m_rest(::std::forward(rest...)...)

我使用自己的 Tuple 实现

//! Declaration of tuple typename with multiple elements
template<typename T, typename... Ts>
class Tuple<T, Ts...>
{
public:
    T m_element;
    Tuple<Ts...> m_rest;

    template<typename U,
             typename... Us,
             typename = class ::std::enable_if<!::std::is_base_of<Tuple,typename ::std::decay<U>::type>::value>::type>
    Tuple(U&& u, Us&&... rest) : m_element(::std::forward<U>(u)), m_rest(::std::forward<Us>(rest)...)
    {
    }
};

template<typename... Ts>
Tuple<typename ::std::decay<Ts>::type...> make_tuple(Ts&&... elements)
{
    return Tuple<typename ::std::decay<Ts>::type...>(::std::forward<Ts>(elements)...);
}

clang 与 GCC 有何不同？我该如何解决这个问题？

谢谢!

Answer 1

What is clang doing differently than GCC? And how can I fix this?

不确定谁是对的，也不确定这能解决您的问题(没有一个最小但完整的问题示例，我无法检查它)但是我有一个错误(仅限 clang++)，当我更改 类时该错误消失了 带有 typename

所以我建议

template<typename U,
         typename... Us,
         // ........VVVVVVVV  <--- "typename" here, not "class"
         typename = typename ::std::enable_if<!::std::is_base_of<Tuple,typename ::std::decay<U>::type>::value>::type>
Tuple(U&& u, Us&&... rest) : m_element(::std::forward<U>(u)), 
                             m_rest(::std::forward<Us>(rest)...)
 { }