第12周项目一（2）-实现复数中的运算符重载

*   *烟台大学计算机学院   *文件名称：xiangmu1.cpp   *作    者：李亚辉   *完成日期：2016年6月21日   *版 本 号：vc++6.0  *   *问题描述： 定义一个完整的类，定义运算符功能，使之能与double型数据进行运算 *输入描述： 无*程序输出： 无.*/  #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 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;};//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; //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+(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)iComplex 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<<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;}//学习心得：进一步学习并运用了运算符重载，感觉在运用友元函数，与不用友元函数时，有很多运用规则、形式发生不同，这是应该注意的点，在基类的基础上拓展其功能，也是对前面一些知识点的总结与运用。还是熟能生巧，感觉自己就是做的太少了，用起来很生涩，虽然老师每一个知识点都会有给的相应的项目进行训练，感觉自己还是做得太少了，果然实践才是硬道理啊。