多继承的解决方法不明确? [英] Ambiguous workaround for multiinheritance?

问题描述

我有一个称为动物的基类，还有一个从动物继承的狗和猫. 还有一个称为dogcat的多继承类，该类继承自dog和cat， 在动物中，我有一种称为睡眠的方法.当我想从Dogcat中使用该方法时，出现错误"DogCat :: sleep"是模棱两可的，我的确理解该问题，但是当我声明睡眠为虚拟时，我在书中读到它应该是可能的-但它不起作用.

I have a base class called animal, and a dog and a cat that inherit from Animal. And a multiinheritance class called dogcat, that inherit from dog and cat, in Animal i have method called sleep. When i want to use that method from dogcat, i get the error "DogCat::sleep" is ambiguous, i do understand the problem, but i read in a book that it should be possible, when you declare sleep as virtual - but it does not work.

这是不可能的吗，这本书是错误的还是有任何解决方法?

Is this not possible is the book wrong or is there any workaround?

class Animal
{
public:
    Animal(){}

    virtual void sleep()
    {
        cout << "zzzzzzzzz" << endl;
    }
    virtual void eat() = 0;

};

class Dog: public Animal
{
protected:
    Dog(){}

    virtual void eat() override
    {
        cout << "eats dogfood" << endl;
    } 
};

class Cat :public Animal
{
public:
    Cat(){}
    virtual void eat() override
    {
        cout << "eats catfood" << endl;
    }
};

class DogCat : public Dog, public Cat
{
public:
    DogCat(){}
    using Dog::eat;

};

int main(int argc, char** argv) {
    DogCat *DC = new DogCat();
    DC->sleep();//Error
}

推荐答案

您遇到了钻石问题

钻石问题"(有时称为致命的死亡钻石" [4])是当两个B类和C类从A继承而D类从B和C继承时产生的歧义.在A中有一个方法B和C已被覆盖，而D没有覆盖它，那么D继承了该方法的哪个版本:B的方法还是C的方法?

The "diamond problem" (sometimes referred to as the "deadly diamond of death"[4]) is an ambiguity that arises when two classes B and C inherit from A, and class D inherits from both B and C. If there is a method in A that B and C have overridden, and D does not override it, then which version of the method does D inherit: that of B, or that of C?

所以.现在您有两个A的实例.解决方案是什么?您有两个:

So. Now you have two instances of A. What is it the solution? You have two:

1. 在一个小节中定义睡眠操作，并将其命名为:

class Cat :public Animal
{
    public:
        Cat(){}
        virtual void eat() override
        {
            cout << "eats catfood" << endl;
        }

        void sleep()
        {
            Animal::sleep();
        }
};

int main(int argc, char** argv) {
    DogCat *DC = new DogCat();
    DC->Cat::sleep();
}

1. 使用虚拟继承，例如@Asesh答案.问题是常见的方法eat().您必须覆盖它.

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