将派生类传递给基类参数的函数

Posted 2021-05-01

我有这个代码：

#include <iostream>

class Base {
  public:
  virtual void sayHello() {
    std::cout << "Hello world, I am Base" << std::endl;
  }
};

class Derived: public Base {
  public:
  void sayHello() {
    std::cout << "Hello world, I am Derived" << std::endl;
  }
};

void testPointer(Base *obj) {
  obj->sayHello();
}

void testReference(Base &obj) {
  obj.sayHello();
}

void testObject(Base obj) {
  obj.sayHello();
}

int main() {
  {
    std::cout << "Testing with pointer argument: ";
    Derived *derived = new Derived;
    testPointer(derived);
  }
  {
    std::cout << "Testing with reference argument: ";
    Derived derived;
    testReference(derived);
  }
  {
    std::cout << "Testing with object argument: ";
    Derived derived;
    testObject(derived);
  }
}

输出是：

Testing with pointer argument: Hello world, I am Derived
Testing with reference argument: Hello world, I am Derived
Testing with object argument: Hello world, I am Base

我的问题是为什么指针案例void testPointer(Base *obj)和引用案例void testReference(Base &obj)都返回void sayHello()的派生实例的结果但是没有和复制案例的传递？我该怎么做才能使复制案例返回派生类函数void sayHello()的结果？