使用enable_if进行C++模板重载:使用g ++和clang进行不同的行为

在解析基类的模板化成员函数的重载期间,我观察到g ++(5.2.1-23)和clang(3.8.0)之间的不同行为-std=c++14.

#include 
#include 

struct Base
{
  template 
  auto a(T t) -> void {
    std::cout<< "False\n";
  }
};

template 
struct Derived : public Base
{

  using Base::a;
  template 
  auto a(T t) -> std::enable_if_t
  {
    std::cout<< "True\n";
  }
};

int main()
{
  Derived d;
  d.a(1); // fails with g++, prints "true" with clang
  Derived d2;
  d2.a(1); // fails with clang++, prints "false" with g++
}

Derived::a使用g ++ 调用失败并显示以下消息:

test.cc: In function ‘int main()’:
test.cc:28:8: error: call of overloaded ‘a(int)’ is ambiguous
   d.a(1);
        ^
test.cc:18:8: note: candidate: std::enable_if_t Derived::a(T) [with T = int; bool B = true; bool Bool = true; std::enable_if_t = void]
   auto a(T t) -> std::enable_if_t
        ^
test.cc:7:8: note: candidate: void Base::a(T) [with T = int]
   auto a(T t) -> void {
        ^

并Derived::a使用以下消息使用clang ++ 调用失败:

test.cc:32:6: error: no matching member function for call to 'a'
  d2.a(1);
  ~~~^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.2.1/../../../../include/c++/5.2.1/type_traits:2388:44: note: candidate template ignored: disabled by 'enable_if' [with T = int, B = false]
    using enable_if_t = typename enable_if<_Cond, _Tp>::type;
                                           ^

我的猜测是,他们对诠释的解释不同using Base::a;,而且在俚语中不予考虑,而在g ++中则考虑(可能太多).我认为会发生的是,如果Derived有true参数,则调用a()调度到Derived的实现,而如果参数是false,则调度调度Base::a.

他们都错了吗？谁是对的？我应该向谁提交错误报告？有人可以解释发生了什么吗？

谢谢