Question

#include <iostream>
using namespace std;

class Criminal {
public:
    virtual void getType() = 0;
    virtual void getCrime() { cout << "Unknown" << endl; }
};

class Thug : public Criminal {
public:
    void getType() { cout << "Thugging" << endl; }
    void getCrime() { cout << "Gangsterism" << endl; }
};

class Tupac : public Thug {
public:
    void getType() { cout << "Rapper" << endl; }
    void getCrime() {
        cout << "Being the best rapper to ever live" << endl;
    }
};

int main() {
    Criminal* tupac = new Tupac();
    Criminal* thug = new Thug();
    Thug* poser = new Tupac(); // Thug has no virtual function
    //Criminal shouldNotCompile;

    tupac->getType();       
    tupac->getCrime();      

    thug->getType();        
    thug->getCrime();       

    poser->getType();       // I intend to call Thug::getType()
    poser->getCrime();      // I intend to call Thug::getCrime()

    delete tupac;
    delete thug;
    delete poser;

    getchar();

    return 0;
}

输出

Rapper
Being the best rapper to ever live
Thugging
Gangsterism
Rapper
Being the best rapper to ever live

但是我打算从暴徒的指针中打电话来打印“Thugging”和“Gangsterism”。

我该怎么做？我希望我的代码按原样运行，因为“Thug”函数不是虚函数，所以不应该从Thug *指针调用任何调用Thug函数的东西吗？

为什么我的代码不按照我的预期方式工作？我的困惑在哪里？

什么是实现我的预期行为的简单方法？

Answer 1

virtual - 继承成员函数。您可能没有将Thug::getType()声明为virtual，但仍然是因为Criminal::getType()。在对象继承自getType()的任何类型上调用Criminal仍将通过虚拟调度。除非您明确指定 getType()所需的内容：

poser->getType(); // virtual dispatch, ends up invoking Tupac::getType()
poser->Thug::getType(); // explicitly call Thug::getType(), no dispatch

这些电话：

delete tupac;
delete thug;
delete poser;

由于Criminal的析构函数不是virtual，

很危险。你实际上没有释放所有内存或摧毁所有成员。

来自基指针的C ++虚拟类？

1 个答案: