C++的运算符重载（二）加号、减号、复制、加等于、等于、不等于、负号、左移、右移、前置++、后置++等的重载

本是古典何须时尚 2023-02-20 12:17 7阅读 0赞

关于运算符重载的相关介绍和注意事项在上一篇博客上已经讲解了，这里就不再赘述了。

*话不多说，直接上代码*

#include <iostream>
    using namespace std;
    
    class Point { 
    private:
    	int m_x;
    	int m_y;
    public:
    	Point(){ }
    	Point(int x,int y):m_x(x),m_y(y){ }
    	void Display(){ 
    		cout << "(" << m_x << "," << m_y << ")" <<endl;
    	}
    
    	Point(const Point &point){ 
    		m_x=point.m_x;
    		m_y=point.m_y;
    	}
    
    	 //加号重载
    	const Point operator+(const Point &point) const { 
    		return Point(m_x+point.m_x,m_y+point.m_y);
    	}
    
    	// 减号重载
    	const Point operator-(const Point &point) const { 
    		return Point(m_x-point.m_x,m_y-point.m_y);
    	}
    
    	// 等于号重载
    	Point& operator= (Point& point){ 
    		m_x=point.m_x;
    		m_x=point.m_x;
    		return *this;
    	}
    
    	// +=重载
    	Point &operator+=(const Point &point) { 
    		m_x += point.m_x;
    		m_y += point.m_y;
    		return *this;
    	}
    
    	// ==重载
    	bool operator==(const Point &point) const { 
    		return (m_x == point.m_x) && (m_y == point.m_y);
    	}
    
    	// ！=重载
    	bool operator!=(const Point &point) const { 
    		return (m_x != point.m_x) || (m_y != point.m_y);
    	}
    
    	// 负号重载
    	const Point operator-() const { 
    		return Point(-m_x, -m_y);
    	}
    
    	// 前置++
    	Point &operator++(){ 
    		m_x++;
    		m_y++;
    		return *this;
    	}
    	
    	// 后置++
    	const Point &operator++(int){ 
    		Point old(m_x,m_y);
    		m_x++;
    		m_y++;
    		return old;
    	}
    
    	friend ostream &operator << (ostream &,const Point &);
    	friend istream &operator >> (istream &cin, Point &point) ;
    };
    
    ostream &operator << (ostream &cout,const Point &point) { 
    	cout << "(" << point.m_x << "," << point.m_y << ")"<< endl ;
    	return cout;
    }
    
    istream &operator >> (istream &cin, Point &point) { 
    	cin >> point.m_x;
    	cin >> point.m_y;
    	return cin;
    }
    
    int main(){ 
    	Point p0(10,20);
    	Point p1(10,10);
    	Point p2(10,20);
    	Point p3(30,30);
    
    	cout << -p0;
    
    	Point p4=p1;
    	p4.Display();
    
    	Point p5=p0+=p1;
    	p5.Display();
    
    	Point p6=p1+p2+p3;
    	p6.Display();
    
    	Point p7=p1+p2-p3;
    	p7.Display();
    
    	cout << (p0==p1) << endl; //p0==p1,返回1，否则返回0
    	cout << (p0!=p1) << endl; //p0!=p1,返回1，否则返回0
    
    	Point p10(100,100);
    	Point p11(200,200);
    	cout << p10++ + p11;
    	cout << ++p10+ p11;
    
    	Point p8;
    	cin >> p8;//从键盘输入
    	cout << p8;
    
    	return 0;
    }

运行结果

(-10,-20)
    (10,10)
    (20,30)
    (50,60)
    (-10,0)
    0
    1
    (300,300)
    (302,302)
    1000  1000 //从键盘输入的
    (1000,1000)

=====================================================================  
**下面讲一些程序中的小细节**

const Point operator+(const Point &point) const { 
    		return Point(m_x+point.m_x,m_y+point.m_y);
    	}

*  第一个 const :  
    我们都知道，两个数相加之后是不可以被赋值的，所以这个const是保证我的返回值是一个const对象，不能再对这个返回值赋值
 *  最后一个const :  
    保证可以执行连续相加的操作（p1+p2+p3），在前面的博客中提到过，const对象是不能调用非const函数的，如果不加const ，p1+p1的返回值就不能够调用operator+函数了

Point &operator+=(const Point &point) { 
    		m_x += point.m_x;
    		m_y += point.m_y;
    		return *this;
    	}

*  这里为什么要返回\*this  
    首先这个函数返回的是一个引用，也就是把它当成对象来使用，返回引用不就是返回对象，返回 \*this不就是返回this所指向的内容。至于这里为什么要用引用？防止产生中间对象

--------------------

*再分享一个运算重载（用户输入两个复数，输出两个复数相加减的结果）*

class Complex{ 
    private:
    	int r,v;
    public:
    	Complex(){ }
    	Complex (int r,int v):r(r),v(v){ }
    
    	Complex& operator+(Complex& c){ 
    		int r=this->r+c.r;
    		int v=this->v+c.v;
    		static Complex ct(r,v);
    		return ct;
    	}
    
    	//minus
    	friend Complex& operator-(const Complex&,const Complex&);
    
    	//output
    	friend ostream& operator << (ostream& ,const Complex&);
    
    	//input
    	friend istream& operator >> (istream& , Complex&);
    };
    
    
    Complex& operator-(const Complex& c1,const Complex& c2){ 
    	int r=c1.r-c2.r;
    	int v=c1.v-c2.v;
    	static Complex ct(r,v);
    	return ct;
    }
    
    ostream& operator<<(ostream& os,const Complex& c){ 
    	int r=c.r;
    	int v=c.v;
    	if(r==0){ 
    		os << v << " ";
    	}else if(v==0){ 
    		os << r;
    	}else if (v<0){ 
    		os << r << v << "i";
    	}else { 
    		os << r << "+" << v <<"i";
    	}
    	return os;
    }
    
    istream& operator >> (istream& is,  Complex& c){ 
    	int r,v;
    	scanf("%d%di",&r,&v);
    	getchar();
    	c.r=r;
    	c.v=v;
    	return is;
    }
    
    int main(){ 
    	Complex c1,c2;
    	cout << "input two number:" << endl;
    
    	cin >> c1 >> c2;
    	cout << "=========" <<endl;
    	cout <<"c1+c2=" << (c1+c2) <<endl;
    	cout <<"c1-c2=" << (c1-c2) <<endl;
    
    	return 0;
    }