将cppleaner的注释转换为答案：

Turning cppleaner's comment into an answer:

来自 namespace.udecl＃15.sentence-1 ：

当using-declarator将基类的声明带入派生类时，派生类中的成员函数和成员函数模板将覆盖和/或隐藏成员函数和成员基类（而不是冲突）中具有相同名称，parameter-type-list，cv-qualification和ref-qualifier（如果有）的功能模板

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，模板参数不计算在内，并且 f 都具有空的parameter-type-list，不是const，也没有ref限定符。

Unfortunately, template parameter doesn't count and both f has empty parameter-type-list, are not const and no ref-qualifier.

Derived :: f ，因此隐藏了 Base :: f 。

gcc

因此，解决该问题的方法是默认参数（返回类型也不计算在内）：

So the way to fix it is by default argument (returned type doesn't count either):

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) { }
};