c++ - 使用声明对两个完全相同名称的名称查找的影响

Question

基本上，我的问题与名称查找和使用声明(http://en.cppreference.com/w/cpp/language/namespace)有关。
假设我们有以下(绝对是愚蠢的)代码:

class Base { 
public:
    void fun()
    {std::cout << "Base fun" << std::endl;}
};

class Derived : public Base { 
public:
    // Here both names "fun" are visible or not?
    using Base::fun;//let's call this func_1

    void fun() //let's call this func_2
    {
        std::cout << "Derived fun" << std::endl;
    }
};

Derived d;
d.fun(); // This resolves to func_2, why?

因此，我的理解是现在我们应该让两个 名称都可见，然后对于名称查找，应该有一些歧义。但实际上并非如此。是什么原因，或者换句话说，我是否误解了一些概念？

Answer 1

标准对这种情况有特殊规定。

When a using-declaration brings names from a base class into a derived class scope, 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 (8.3.5), cv-qualification, and ref-qualifier (if any) in a base class (rather than conflicting).

([命名空间.udecl]/15)

请注意，与往常一样，您可以通过执行 d.Base::fun() 强制调用 Base::fun。