C++多继承静态函数调用歧义

Question

我从两个不同的基类派生一个类，这两个基类都有一个同名的静态函数。

为了解决这种歧义，我尝试使用范围运算符 - 就像我对成员函数所做的那样。但是，这不会编译。 为什么？语法错误？

我想通过派生类型名而不是直接通过基类名调用静态函数。其实我更想阻止这种情况，但我不知道该怎么做。

当我离开模板时，以下代码中的错误（已注释掉）也会发生：

#include <iostream>

template<class TDerived>
class StaticBaseA
{
public:
    static void announce()
    {
        std::cout << "do something" << std::endl;
    }
};

template<class TDerived>
class StaticBaseB
{
public:
    static void announce()
    {
        std::cout << "do something else" << std::endl;
    }
};

class Derived :
      public StaticBaseA<Derived>
    , public StaticBaseB<Derived>
{
    using StaticBaseA<Derived>::announce;
};

class NonDerived {};

int main(int argc, char* argv[])
{
    Derived::announce();
    // What I want:
    //Derived::StaticBaseB<Derived>::announce(); Error: "Undefined symbol 'StaticBaseB'

    // What works, but what I don't want ...
    StaticBaseB<Derived>::announce();

    // ... because I would like to prevent this (however this is done):
    StaticBaseB<NonDerived>::announce();


    return 0;
}

Answer 1

在StaticBaseA 和StaticBaseB 中“宣布”protected 可能是做你想做的事的一部分。

然后您就无法从 main 调用 StaticBaseB<NonDerived>::announce，因为它将无法访问。您可以从派生自 StaticBaseB 的类中调用它。

换句话说：

template<class TDerived>
class StaticBaseA
{
protected:
   static void announce()
   {
       std::cout << "do something" << std::endl;
   }
};

template<class TDerived>
class StaticBaseB
{
protected:
   static void announce()
   {
       std::cout << "do something else" << std::endl;
    }
};

在 Derived 中，您必须向公众宣传“公告”。

class Derived : public StaticA<Derived>, public StaticB<Derived >
{
  public:
     using StaticA<Derived>::announce;
};

int main()
{
     Derived::announce(); // legal and calls StaticBaseA::announce
     NotDerived::announce(); // no such function
     StaticBaseA< Derived >::announce(); // not accessible
     StaticBaseB< Derived >::announce(); // also not accessible
     StaticBaseA< NotDerived >::announce(); // not accessible
     StaticBaseB< NotDerived >::announce(); // also not accessible
}