第八周作业任务一方案三

方案三：在方案二的基础上，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运算的结果（其中?为+、-、*、/之一）。另外，定义一目运算符-，-c相当于0-c。

 

#include <iostream>   using namespace std;  class Complex  {  public:      Complex(){real=0;imag=0;}      Complex(double r,double i){real=r;imag=i;}      Complex operator-();      friend Complex operator+(Complex &c1, Complex &c2);      friend Complex operator+(double d1, Complex &c2);      friend Complex operator+(Complex &c1, double d2);      friend Complex operator-(Complex &c1, Complex &c2);      friend Complex operator-(double d1, Complex &c2);      friend Complex operator-(Complex &c1, double d2);      friend Complex operator*(Complex &c1, Complex &c2);      friend Complex operator*(double d1, Complex &c2);      friend Complex operator*(Complex &c1, double d2);      friend Complex operator/(Complex &c1, Complex &c2);      friend Complex operator/(double d1, Complex &c2);      friend Complex operator/(Complex &c1, double d2);      void display();  private:      double real;      double imag;  };    Complex Complex::operator-()  {      return(0-*this);  }    //复数相加：(a+bi)+(c+di)=(a+c)+(b+d)i.    Complex operator+(Complex &c1, Complex &c2)  {      Complex c;      c.real=c1.real+c2.real;      c.imag=c1.imag+c2.imag;      return c;  }  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;  }  //复数相减：(a+bi)-(c+di)=(a-c)+(b-d)i.   Complex operator-(Complex &c1, Complex &c2)  {      Complex c;      c.real=c1.real-c2.real;      c.imag=c1.imag-c2.imag;      return c;  }  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;  }    //复数相乘：(a+bi)(c+di)=(ac－bd)+(bc+ad)i.   Complex operator*(Complex &c1, Complex &c2)  {      Complex c;      c.real=c1.real*c2.real-c1.imag*c2.imag;      c.imag=c1.imag*c2.real+c1.real*c2.imag;      return c;  }  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;  }    //复数相除：(a+bi)/(c+di)=(ac+bd)/(c^2+d^2) +(bc-ad)/(c^2+d^2)i    Complex operator/(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;  }  Complex operator/(Complex &c1, double d2)  {      Complex c(d2,0);      return c1/c;  }    void Complex::display()  {      cout<<"("<<real<<","<<imag<<"i)"<<endl;  }    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;      cout<<"-c1=";(-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();      c3=c1/c2;      cout<<"c1/c2=";   c3.display();      cout<<"c1/d=";    (c1/d).display();      cout<<"d/c1=";    (d/c1).display();        system("pause");      return 0;  }  

上机感言：对double型数据的运用还是一知半解、知识综合起来运用还是吃力一些。。。