首页 > 解决方案 > 从基类调用派生类中的非虚函数(C++)

问题描述

test不是虚函数,在基类和派生类中都定义了。我有一个派生类对象。它最终调用test基类而不是派生类中的基类。为什么会这样?如何设置它以便根据对象类型调用测试。

电流输出

B func                                                                                                         
A f1                                                                                                           
A test

我希望它是

B func                                                                                                         
A f1                                                                                                           
B test
#include <iostream>

using namespace std;
class A {
    protected:
    void test()
    {
        cout<<"A test"<<endl;
    }
    void f1(){
        cout<<"A f1" <<endl;
        test();
    }
};
class B: public A {
    protected: 
    void test()
    {
        cout<<"B test" <<endl;
    }
   public:
    void func(){
        cout<<"B func" << endl;
        f1();

    }

};
int main()
{
    B tmp;
    tmp.func();
    return 0;
}

标签: c++inheritance

解决方案


有 2 种方法可以存档所需的结果:

使用将virtual
基类函数声明为虚拟函数将确保默认调用目标函数的最高继承。例如,在您的情况下:

class A {
protected:

    /**
     * Declared also in the derived class, so as long the current object is the derived class,
     * this function will be called from the derived class.
     * [Enabled only due to the use of the `virtual` keyword].
     */
    virtual void test() {
        cout << "A test" << endl;
    }

    void f1() {
        cout << "A f1" << endl;
        test();
    }
};

使用 CRTP [巨大的开销来存档可以使用虚拟存档的内容]

template <class Derived> // Template param that will accept the derived class that inherit from this class.
class A {
protected:
    virtual void test() const {
        cout << "A test" << endl;
    }

    void f1() const {
        cout << "A f1" << endl;
        static_cast<Derived const&>(*this).test(); // Call test() function from the derived class.
    }
};

class B: public A<B> { // inherit A that accepts template argument this derived class
// protected:
public: // Pay attention that now test should be public in the derived class in order to access it via the base class.
    void test() const override {
        cout << "B test" << endl;
    }

//public:
    void func() const {
        cout << "B func" << endl;
        f1();
    }
};

阅读更多关于 CRTP的信息。


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