第八週作業任務一方案三

方案三:在方案二的基礎上,擴展+、-、*、/運算符的功能,使之能與double型數據進行運算。設Complex c; double d; c?d和d?c的結果爲將d視爲實部爲d的複數同c運算的結果(其中?爲+、-、*、/之一)。另外,定義一目運算符-,-c相當於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);  
}  
  
//複數相加:(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;  
}  
//複數相減:(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;  
}  
  
//複數相乘:(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;  
}  
  
//複數相除:(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;  
}  


上機感言:對double型數據的運用還是一知半解、知識綜合起來運用還是吃力一些。。。

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