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C++運(yùn)算符重載成員函數(shù)與友元函數(shù)詳解

更新時(shí)間：2013年07月31日 09:47:48 作者：

以下是對(duì)C++運(yùn)算符重載成員函數(shù)與友元函數(shù)進(jìn)行了介紹，需要的朋友可以過(guò)來(lái)參考下

#include<iostream>
using namespace std;
class A
{
    int x,y;
    public:
    A(int xx,int yy):x(xx),y(yy){}
    A(){x=0;y=0;}
    A operator+(const A&b) //不加const限定，也可以
    { return A(x+b.x,y+b.y); }
    A operator-()
    { return A(-x,-y); }
    void show()
    {cout<<"x="<<x<<" y="<<y<<endl;}
};
void test_A()
{
    A a1(2008,512),a2(2013,420),a3;
    a3=a1+a2; //調(diào)用操作符重載函數(shù): a1.oprator +(a2)
    a3.show();
    a1=-a1; //調(diào)用操作符重載函數(shù): a1.operator -()
    a1.show();
}
/***********************
執(zhí)行結(jié)果
x=4021 y=93
x=-2008 y=-512
**********************/
class B
{
    int x,y;
    public:
        B(int xx,int yy):x(xx),y(yy){}
        B(){x=0;y=0;}
        friend B operator+(const B&a,const B&b);
        friend B operator-(const B&a);
        void show()
        {cout<<"x="<<x<<" y="<<y<<endl;};
};
B operator+(const B&a,const B&b)
{return B(a.x+b.x,a.y+b.y);}
B operator-(const B&a)
{return B(-a.x,-a.y);}
/***************************
class B
{
    int x,y;
    public:
        B(int xx,int yy):x(xx),y(yy){}
        B(){x=0;y=0;}
        friend B operator+(const B&a,const B&b)
                {return B(a.x+b.x,a.y+b.y);}
        friend B operator-(const B&a)
            {return B(-a.x,-a.y);}
        void show()
        {cout<<"x="<<x<<" y="<<y<<endl;};
}
********************************/
int main()
{
    B B1(1991,1105),B2(2013,62),B3;
    B3=B1+B2; //調(diào)用操作符重載函數(shù): a1.oprator +(a2)
    B3.show();
    B1=-B1; //調(diào)用操作符重載函數(shù): a1.operator +()
    B1.show();
}
/****************************
運(yùn)行結(jié)果：
x=4004 y=1167
x=-1991 y=-1105
Process returned 0 (0x0)   execution time : 0.021 s
Press any key to continue.

*****************************/

復(fù)制代碼代碼如下:

#include<iostream>
using namespace std;
class A
{
    int x,y;
    public:
    A(int xx,int yy):x(xx),y(yy){}
    A(){x=0;y=0;}
    A operator+(const A&b) //不加const限定，也可以
    { return A(x+b.x,y+b.y); }
    A operator-()
    { return A(-x,-y); }
    void show()
    {cout<<"x="<<x<<" y="<<y<<endl;}
};
void test_A()
{
    A a1(2008,512),a2(2013,420),a3;
    a3=a1+a2; //調(diào)用操作符重載函數(shù): a1.oprator +(a2)
    a3.show();
    a1=-a1; //調(diào)用操作符重載函數(shù): a1.operator -()
    a1.show();
}
/***********************
執(zhí)行結(jié)果
x=4021 y=93
x=-2008 y=-512
**********************/
class B
{
    int x,y;
    public:
        B(int xx,int yy):x(xx),y(yy){}
        B(){x=0;y=0;}
        friend B operator+(const B&a,const B&b);
        friend B operator-(const B&a);
        void show()
        {cout<<"x="<<x<<" y="<<y<<endl;};
};
B operator+(const B&a,const B&b)
{return B(a.x+b.x,a.y+b.y);}
B operator-(const B&a)
{return B(-a.x,-a.y);}
/***************************
class B
{
    int x,y;
    public:
        B(int xx,int yy):x(xx),y(yy){}
        B(){x=0;y=0;}
        friend B operator+(const B&a,const B&b)
                {return B(a.x+b.x,a.y+b.y);}
        friend B operator-(const B&a)
            {return B(-a.x,-a.y);}
        void show()
        {cout<<"x="<<x<<" y="<<y<<endl;};
}
********************************/
int main()
{
    B B1(1991,1105),B2(2013,62),B3;
    B3=B1+B2; //調(diào)用操作符重載函數(shù): a1.oprator +(a2)
    B3.show();
    B1=-B1; //調(diào)用操作符重載函數(shù): a1.operator +()
    B1.show();
}
/****************************
運(yùn)行結(jié)果：
x=4004 y=1167
x=-1991 y=-1105
Process returned 0 (0x0)   execution time : 0.021 s
Press any key to continue.
*****************************/