第八周项目一 实现复数类中的运算符重载-(3)扩展(2)中的+-*/功能

来源:互联网 发布:安知玉如意说的是什么 编辑:程序博客网 时间:2024/06/05 05:11

项目要求

定义一个定义完整的类(是可以当作独立的产品发布,成为众多项目中的“基础工程”)。这样的类在(2)的基础上,扩展+、-、*、/运算符的功能,使之能与double型数据进行运算。设Complex c; double d; c+d和d+c的结果为“将d视为实部为d的复数同c相加”,其他-、*、/运算符类似。


代码如下

#include <iostream>using namespace std;class Complex{public:    Complex(){real=0;imag=0;}    Complex(double r,double i){real=r; imag=i;}    friend Complex operator+(Complex &c1,Complex &c2);    friend Complex operator+(double d,Complex &c1);    friend Complex operator+(Complex &c1,double d);    friend Complex operator-(Complex &c1,Complex &c2);    friend Complex operator-(double d,Complex &c1);    friend Complex operator-(Complex &c1,double d);    friend Complex operator*(Complex &c1,Complex &c2);    friend Complex operator*(double d,Complex &c1);    friend Complex operator*(Complex &c1,double d);    friend Complex operator/(Complex &c1,Complex &c2);    friend Complex operator/(double d,Complex &c1);    friend Complex operator/(Complex &c1,double d);    void display();private:    double real;    double imag;};//下面定义友元函数//加法Complex operator+(Complex &c1,Complex &c2){    return Complex(c1.real+c2.real,c1.imag+c2.imag);}Complex operator+(double d,Complex &c1){    Complex c(d,0);    return c+c1;}Complex operator+(Complex &c1,double d){    Complex c(d,0);    return c+c1;}//减法Complex operator-(Complex &c1,Complex &c2){    return Complex(c1.real-c2.real,c1.imag-c2.imag);}Complex operator-(double d,Complex &c1){    Complex c(d,0);    return c-c1;}Complex operator-(Complex &c1,double d){    Complex c(d,0);    return c1-c;}//乘法Complex operator*(Complex &c1,Complex &c2){    return Complex(c1.real*c2.real,c1.imag*c2.imag);}Complex operator*(double d,Complex &c1){    Complex c(d,0);    return c*c1;}Complex operator*(Complex &c1,double d){    Complex c(d,0);    return c*c1;}//除法Complex operator/(Complex &c1,Complex &c2){    return Complex(c1.real/c2.real,c1.imag/c2.imag);}Complex operator/(double d,Complex &c1){    Complex c(d,0);    return c/c1;}Complex operator/(Complex &c1,double d){    Complex c(d,0);    return c1/c;}void Complex::display(){    cout<<"("<<real<<","<<imag<<")"<<endl;}//下面定义用于测试的main()函数//main函数偷的贺老师的~int main(){    Complex c1(3,4),c2(5,-10),c3;    double d=11;    cout<<"c1=";    c1.display();    cout<<"c2=";    c2.display();    cout<<"d="<<d<<endl<<endl;    cout<<"下面是重载运算符的计算结果: "<<endl;    c3=c1+c2;    cout<<"c1+c2=";    c3.display();    cout<<"c1+d=";    (c1+d).display();    cout<<"d+c1=";    (d+c1).display();    c3=c1-c2;    cout<<"c1-c2=";    c3.display();    cout<<"c1-d=";    (c1-d).display();    cout<<"d-c1=";    (d-c1).display();    c3=c1*c2;    cout<<"c1*c2=";    c3.display();    cout<<"c1*d=";    (c1*d).display();    cout<<"d*c1=";    (d*c1).display();    c3=c1/c2;    cout<<"c1/c2=";    c3.display();    cout<<"c1/d=";    (c1/d).display();    cout<<"d/c1=";    (d/c1).display();    return 0;}


运行结果



学习心得

在定义+-*/时写上这步Complex c(d,0);并且return c*c1;很有必要。不仅可以用上之前构造的函数,而且显得很机智。

直接写成    c1.real*=d;再return c1 ;也是可以的


0 0
原创粉丝点击