第十二周项目一（3）-实现复数类中的运算符重载

问题及代码：

/*
*烟台大学计算机学院
*文件名称：xiangmu1(3).cpp
*作    者：闫安
*完成日期：2016年5月30日
*版 本 号：codeblocks 16.01
*
*问题描述：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。
*程序输入：无
*程序输出：运算结果
*/
#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+(const Complex &c1,Complex &c2);
    friend Complex operator-(const Complex &c1,Complex &c2);
    friend Complex operator*(const Complex &c1,Complex &c2);
    friend Complex operator/(const Complex &c1,Complex &c2);
    friend Complex operator+(double d1, Complex &c2);
    friend Complex operator-(double d1, Complex &c2);
    friend Complex operator*(double d1, Complex &c2);
    friend Complex operator/(double d1, Complex &c2);
    friend Complex operator+(Complex &c1, double d2);
    friend Complex operator-(Complex &c1, double d2);
    friend Complex operator*(Complex &c1, double d2);
    friend Complex operator/(Complex &c1, double d2);
    void display();
private:
    double real;
    double imag;
};
//下面定义成员函数
void Complex::display()
{
    cout<<"("<<real<<","<<imag<<"i)"<<endl;
}
Complex operator+(const Complex &c1,Complex &c2)
{
    Complex c;
    c.real=c1.real+c2.real;
    c.imag=c1.imag+c2.imag;
    return c;
}
Complex operator-(const Complex &c1,Complex &c2)
{
    Complex c;
    c.real=c1.real-c2.real;
    c.imag=c1.imag-c2.imag;
    return c;
}
Complex operator*(const Complex &c1,Complex &c2)
{
    Complex c;
    c.real=c1.real*c2.real;
    c.imag=c1.imag*c2.imag;
    return c;
}
//复数相除：(a+bi)/(c+di)=(ac+bd)/(c^2+d^2) +(bc-ad)/(c^2+d^2)i
Complex operator/(const Complex &c1,Complex &c2)
{
   Complex c;  
    c.real=(c1.real*c2.real+c1.imag*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag);  
    c.imag=(c1.imag*c2.real-c1.real*c2.imag)/(c2.real*c2.real+c2.imag*c2.imag);  
    return c;  
}
Complex operator+(double d1, Complex &c2)
{
    Complex c(d1,0);
    return c+c2; //按运算法则计算的确可以，但充分利用已经定义好的代码，既省人力，也避免引入新的错误，但可能机器的效率会不佳

}
Complex operator-(double d1, Complex &c2)
{
    Complex c(d1,0);
    return c-c2;
}
Complex operator*(double d1, Complex &c2)
{
    Complex c(d1,0);
    return c*c2;
}
Complex operator/(double d1, Complex &c2)
{
    Complex c(d1,0);
    return c/c2;
}
Complex operator+(Complex &c1, double d2)
{
    Complex c(d2,0);
    return c1+c;
}
Complex operator-(Complex &c1, double d2)
{
    Complex c(d2,0);
    return c1-c;
}
Complex operator*(Complex &c1, double d2)
{
    Complex c(d2,0);
    return c1*c;
}
Complex operator/(Complex &c1, double d2)
{
    Complex c(d2,0);
    return c1/c;
}
//下面定义用于测试的main()函数
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：