C++ 在子类中创建一个可用的方法

Question

这是我的基类

class Base {
protected:
    int number;
public:
    Base(int num);
    virtual void display() = 0;  
};

这两个类都继承了Base。

class Derived: public Base {
public:
    Derived(int num);
    void display(){cout <<"hello";}
    void other();
};

class Derived2: public Base {
public:
    Derived(int num);
    void display(){cout <<"hello other";}
};

这个类允许我实例化我的两个类 Derived 和 Derived2

class Foo{
private:
    Base * toto[2];
public:
    Foo(){
     toto[0] = new Derived;
     toto[1] = new Derived2;
    }

    void doSomething();
}

我想做这个

void Foo::doSomething(){
    toto[0]->other();
}

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最后一个问题我创建了一个继承了Derived2和Base的新类

class Derived3: public Derived2, public Base {
public:
    Derived(int num);
    void display(){cout <<"hello";}
    void other();
};

现在我想做这个

toto[0] = new Derived3

This my error message 

从派生类“Derived3”到基类“Base”的转换不明确: Derived3 类 -> Base 类 Derived3 类 -> Derived2 类 -> Base 类

从不兼容类型“Derived3 *”分配给“Base *”

Answer 1

如果 toto[0] 总是指向 Derived 而 toto[1] 总是指向 Derived2，你应该这样做:

class Foo{
private:
    Derived d1;
    Derived2 d2;
    Base * toto[2];
public:
    Foo(){
        toto[0] = &d1;
        toto[1] = &d2;
    }

    void doSomething() {
        d1.other();
    }
};

如果您确实有诚意需要创建无法静态检查键入的情况，请使用dynamic_cast。但如果可能的话，尽量避免这种情况。不要将 C++ 视为动态类型语言。

Derived* dptr = dynamic_cast<Derived*>(toto[0]);
if (dptr) {
    dptr->other();
} else {
    throw std::logic_error("toto[0] should point to Derived");
}