c++继承体系中的拷贝控制成员的定义

最新推荐文章于 2025-01-20 01:14:44 发布

与狼共舞-C

最新推荐文章于 2025-01-20 01:14:44 发布

阅读量854

点赞数 1

CC 4.0 BY-SA版权

分类专栏： C++

本文链接：https://blog.youkuaiyun.com/u011402017/article/details/52789747

C++ 专栏收录该内容

13 篇文章

订阅专栏

本文通过具体的C++代码示例，详细介绍了基类和派生类中拷贝构造函数及拷贝赋值运算符的实现方式。同时，展示了移动构造函数和移动赋值运算符的具体应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/***验证继承体系中的拷贝控制成员***/
#include<iostream>
using namespace std;
class Base {
public:
	Base() = default;
	Base(int i, int j) :num1(i), num2(j) { cout << "Base  construct" << endl; }
	Base(const Base& rhs) :num1(rhs.num1), num2(rhs.num2) { cout << "Base copy construct " << endl; }
	Base& operator=(const Base& rhs)
	{
		num1 = rhs.num1;
		num2 = rhs.num2;
		cout << "Base copy =" << endl;
		return *this;
	}
	Base(Base&& rhs):num1(rhs.num1), num2(rhs.num2) { cout << "Base move construct " << endl; }
	Base& operator=(Base&& rhs)
	{
		num1 = rhs.num1;
		num2 = rhs.num2;
		cout << "Base move =" << endl;
		return *this;
	}
	virtual ~Base() { cout << "Base virtual destroy function" << endl; }
	int num1;
	int num2;

};

class Dervied :public Base {
public:
	Dervied() = default;
	Dervied(int i , int j, int m , int n) :Base(i, j), num3(m), num4(n) { cout << "Derived  construct" << endl; }
	Dervied(const Dervied& rhs) :Base(rhs), num3(rhs.num3), num4(rhs.num4) { cout << "Dervied copy construct" << endl; }  //拷贝构造函数
	Dervied& operator=(const Dervied& rhs)   //拷贝复制运算符
	{
		Base::operator=(rhs);
		num3 = rhs.num3;
		num4 = rhs.num4;
		cout << "Dervied copy = " << endl;
		return *this;
	}
	Dervied(Dervied&& rhs) :Base(std::move(rhs)), num3(rhs.num3), num4(rhs.num4) { cout << "Dervied move construct" << endl; }//移动构造函数
	Dervied& operator=(Dervied&& rhs)                           //移动赋值运算符
	{
		Base::operator=(std::move(rhs));
		num3 = rhs.num3;
		num4 = rhs.num4;
		cout << "Dervied move =" << endl;
		return *this;
	}

	virtual ~Dervied()
	{
		cout << "Dervied destroy" << endl;
	}
	int num3;
	int num4;
};