本文主要是介绍9.2(C++ Day 4),希望对大家解决编程问题提供一定的参考价值,需要的开发者们随着小编来一起学习吧!
一、作业
完成算术运算符重载,实现至少两个运算符的成员函数和全局函数的版本
1.(1)成员函数实现算术运算符(-)重载
const 类名 operator#(const 类名 &R) const
{}
#include <iostream>using namespace std;class Person
{
private:int a;int b;
public:Person(){}Person(int a, int b):a(a),b(b){}void show(){cout << "a = " << a << endl;cout << "b = " << b << endl;}const Person operator-(const Person &R) const{Person temp;temp.a = a - R.a;temp.b = b - R.b;return temp;}
};
int main()
{Person p1(30,20);Person p2(10,10);Person p3 = p1 - p2;p3.show();return 0;
}
(2)全局函数实现算术运算符(-)重载
const 类名 operator#(const 类名 &L, const 类名 &R)
{}
#include <iostream>using namespace std;class Person
{friend const Person operator-(const Person &L, const Person &R);
private:int a;int b;
public:Person(){}Person(int a, int b):a(a),b(b){}void show(){cout << "a = " << a << endl;cout << "b = " << b << endl;}};
const Person operator-(const Person &L, const Person &R)
{Person temp;temp.a = L.a - R.a;temp.b = L.b - R.b;return temp;
}
int main()
{Person p1(30,20);Person p2(10,10);Person p3 = p1 - p2;p3.show();return 0;
}
实现结果:
2.(1)成员函数实现算术运算符(*)重载
const 类名 operator#(const 类名 &R) const
{}
#include <iostream>using namespace std;class Person
{
private:int a;int b;
public:Person(){}Person(int a, int b):a(a),b(b){}void show(){cout << "a = " << a << endl;cout << "b = " << b << endl;}const Person operator*(const Person &R) const{Person temp;temp.a = a * R.a;temp.b = b * R.b;return temp;}
};
int main()
{Person p1(3,2);Person p2(2,4);Person p3 = p1 * p2;p3.show();return 0;
}
(2)全局函数实现算术运算符(*)重载
const 类名 operator#(const 类名 &L, const 类名 &R)
{}
#include <iostream>using namespace std;class Person
{friend const Person operator*(const Person &L, const Person &R);
private:int a;int b;
public:Person(){}Person(int a, int b):a(a),b(b){}void show(){cout << "a = " << a << endl;cout << "b = " << b << endl;}};
const Person operator*(const Person &L, const Person &R)
{Person temp;temp.a = L.a * R.a;temp.b = L.b * R.b;return temp;
}
int main()
{Person p1(3,20);Person p2(10,10);Person p3 = p1 * p2;p3.show();return 0;
}
代码实现:
二、思维导图
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