Question

我有这段代码：

#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);
  }
}

输出结果为：

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返回复制案例的传递？我该怎么做才能使副本案例返回派生类函数void sayHello()的结果？

Answer 1

引用引用或指针的函数引用传入的原始对象，而按值参数将创建对象的副本。因为你只是复制基础部分（因为它需要一个基础对象），你最终只使用基础部分的副本，它就像一个基础，因为它是一个基础。 / p>

这个＆＃34; base-only＆＃34;复制被称为＆＃34;切片＆＃34;因为它只复制你的部分对象，＆＃34;切掉＆＃34;派生的部分。

将派生类传递给基类参数的函数

1 个答案: