c++ - 重写非常量虚方法是否隐藏了常量重载？

Question

考虑:

#include <iostream>

using namespace std;

struct A {
  virtual void f() { cout << "A::f" << endl; }
  virtual void f() const { cout << "A::f const" << endl; }
};

struct B : public A {};

struct C : public A {
   virtual void f() { cout << "C::f" << endl; }
};


int main()
{
   const B b;
   b.f();   // prints "A::f const"

   const C c;
   c.f();
   // Compile-time error: passing ‘const C’ as ‘this’ argument of
   //   ‘virtual void C::f()’ discards qualifiers
}

(我正在使用 GCC。)

所以 f() 的 const 版本似乎隐藏在 C 中。这对我来说很有意义，但这是标准强制要求的吗？

Answer 1

我将(再次)链接这个很棒的 article :

First, [the compiler] looks in the immediate scope, in this case the scope of class C, and makes a list of all functions it can find that are named f (regardless of whether they're accessible or even take the right number of parameters). Only if it doesn't does it then continue "outward" into the next enclosing scope [...]

是的，f 的const 版本是隐藏的，这是完全正常的。正如 Simone 所指出的，您可以使用 using 语句将 A::f 引入 C 范围。