基础和派生模板成员之间的重载解析 类

Question

Microsoft 编译器 (Visual Studio 2017 15.2) 拒绝以下代码：

#include <type_traits>

struct B
{ 
    template<int n, std::enable_if_t<n == 0, int> = 0>
    void f() { }
};

struct D : B
{
    using B::f; 
    template<int n, std::enable_if_t<n == 1, int> = 0>
    void f() { }
};

int main()
{
    D d;
    d.f<0>();
    d.f<1>();
}

错误是：

error C2672: 'D::f': no matching overloaded function found
error C2783: 'void D::f(void)': could not deduce template argument for '__formal'
note: see declaration of 'D::f'

Clang 也拒绝了它：

error: no matching member function for call to 'f'
    d.f<0>();
    ~~^~~~
 note: candidate template ignored: disabled by 'enable_if' [with n = 0]
    using enable_if_t = typename enable_if<_Cond, _Tp>::type;

GCC 完全接受它。哪个编译器是正确的？

加法：

与 SFINAE 的形式

template<int n, typename = std::enable_if_t<n == 0>>
...
template<int n, typename = std::enable_if_t<n == 1>>

GCC 也产生错误：

error: no matching function for call to ‘D::f<0>()’
 d.f<0>();
        ^
note: candidate: template<int n, class> void D::f()
 void f()
      ^
note:   template argument deduction/substitution failed:

Answer 1

将 cppleaner 的评论转化为答案：

来自 namespace.udecl#15.sentence-1:

When a using-declarator brings declarations from a base class into a derived class, member functions and member function templates in the derived class override and/or hide member functions and member function templates with the same name, parameter-type-list, cv-qualification, and ref-qualifier (if any) in a base class (rather than conflicting)

不幸的是，模板参数不算在内，并且 f 的参数类型列表为空，不是 const，也没有引用限定符。

Derived::f 所以隐藏 Base::f.

gcc 接受该代码是错误的。

所以解决方法是默认参数（返回类型也不算）：

struct B
{ 
    template <int n>
    void f(std::enable_if_t<n == 0>* = nullptr) { }
};

struct D : B
{
    using B::f; 
    template <int n>
    void f(std::enable_if_t<n == 1>* = nullptr) { }
};