使用基成员函数的重载决议引入派生 class

Question

根据标准中的一些引述：
[over.match.funcs]/4

[...] For non-conversion functions introduced by a using-declaration into a derived class, the function is considered to be a member of the derived class for the purpose of defining the type of the implicit object parameter.

并考虑以下代码：

#include <iostream>
struct Base {
    void show(double) {
        std::cout << "0" << std::endl;
    }
};
struct Test :Base {
    using Base::show;  //#1
    void show(double) { //#2
        std::cout << "1" << std::endl;
    }
};
int main() {
    Test t;
    t.show(1.2);
}

根据我引用的标准，意思是将Test类型的隐式对象参数作为#1类型的隐式对象参数。
对于 #1，声明将是 show(Test&,double).
对于 #2，声明将是 show(Test&,double)。
所以为了重载决议，#1的每个隐式转换序列与#2的隐式转换序列没有区别，t.show(1.2)的调用会产生歧义，但是#2被调用，为什么？如果我在标准中遗漏了什么，请纠正我。

Answer 1

Test::show(double) 与 Base::show(double) 具有相同的签名；它只是 隐藏了 基础 class.

对于using declaration,

（强调我的）

If the derived class already has a member with the same name, parameter list, and qualifications, the derived class member hides or overrides (doesn't conflict with) the member that is introduced from the base class.

根据标准，[namespace.udecl]/14

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 ([dcl.fct]), trailing requires-clause (if any), cv-qualification, and ref-qualifier (if any), in a base class (rather than conflicting). Such hidden or overridden declarations are excluded from the set of declarations introduced by the using-declarator. [ Example:
struct B {
  virtual void f(int);
  virtual void f(char);
  void g(int);
  void h(int);
};

struct D : B {
  using B::f;
  void f(int);      // OK: D::f(int) overrides B::f(int);

  using B::g;
  void g(char);     // OK

  using B::h;
  void h(int);      // OK: D::h(int) hides B::h(int)
};

...

使用基成员函数的重载决议引入派生 class

the overload resolution of using base member function introduced into derived class

c++

using-declaration

language-lawyer