Question

我的情况是我从两个不同的基类派生一个类，它们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。但是这不编译。 为什么？语法错误？

我想通过派生的typename调用静态函数，而不是直接通过基类名称。其实我更愿意预防这种情况，但我不知道该怎么做。

当我离开模板时，下面的代码中也会出现错误（已注释掉）：

#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

制作＆＃34;宣布＆＃34; protected StaticBaseA和StaticBaseB中的StaticBaseB<NonDerived>::announce可能会成为你想做的事情的一部分。

然后你无法从main调用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; } };，因为它无法访问。您可以从派生自StaticBaseB的类中调用它。

换句话说：

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
}

在Derived中你必须宣传＆＃34;宣布＆＃34;公开。

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C ++多重继承静态函数调用歧义

1 个答案: