C++第9周项目1 - 实现复数类中的运算符重载

课程首页地址：http://blog.csdn.net/sxhelijian/article/details/7910565，本周题目链接：http://blog.csdn.net/sxhelijian/article/details/8841620

【项目1-Complex类】接第8周项目1，定义Complex类中的<<和>>运算符的重载，实现输入和输出，改造原程序中对运算结果显示方式，使程序读起来更自然。

参考解答：（在VS2008中调试通过，CodeBlocks中有编译错误，见后说明。）

#include <iostream>using namespace std;class Complex{public:Complex(){real=0;imag=0;}Complex(double r,double i){real=r;imag=i;}    //Complex &operator=(Complex &c){real=c.real;imag=c.imag;return *this;}Complex operator-();//实现输入、输出的运算符重载friend ostream& operator<< (ostream& output, Complex& c);friend istream& operator>> (istream& input,Complex& c);//实现加减乘除的运算符重载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);private:double real;double imag;};//实现输出的运算符重载ostream &operator << (ostream &output, Complex &c){    output<<"("<<c.real;if(c.imag>=0) output<<"+";output<<c.imag<<"i)";return output;}//实现输入的运算符重载istream& operator >> (istream& input,Complex& c){    int a,b;char sign,i;do{cout<<"input a complex number(a+bi或a-bi):";input>>a>>sign>>b>>i;}while(!((sign=='+'||sign=='-')&&i=='i'));c.real=a;c.imag=(sign=='+')?b:-b;return input;}Complex Complex::operator-(){return(0.0-*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 c0(d1,0),c;return c0+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 c0(d2,0),c;c=c1+c0;return 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;}int main(){Complex c1,c2,c3,c;double d=11;cout<<"c1: "<<endl;;cin>>c1;cout<<"c2: "<<endl;cin>>c2;cout<<"c1="<<c1<<endl;cout<<"c2="<<c2<<endl;cout<<"d="<<d<<endl;cout<<"-c1="<<(-c1);c3=c1+c2;cout<<"c1+c2="<<c3<<endl;cout<<"c1+d="<<(c1+d)<<endl;cout<<"d+c1="<<(d+c1)<<endl;c3=c1-c2;cout<<"c1-c2="<<c3<<endl;cout<<"c1-d="<<(c1-d)<<endl;cout<<"d-c1="<<(d-c1)<<endl;c3=c1*c2;cout<<"c1*c2="<<c3<<endl;cout<<"c1*d="<<(c1*d)<<endl;cout<<"d*c1="<<(d*c1)<<endl;c3=c1/c2;cout<<"c1/c2="<<c3<<endl;cout<<"c1/d="<<(c1/d)<<endl;cout<<"d/c1="<<(d/c1)<<endl;return 0;}

　　CodeBlocks中有编译错误发生在第149、152、153、156、157、160、161、164、165行。146、147、151等直接输出复数类对象没有出问题，而问题出在了输出重载的运算结果的输出。这说明不是operator<<()重载有问题，如果将第149行写为：

cout<<"-c1=";Complex c=-c1;cout<<c;

　　问题解决，其他地方类似。

　　曾经怀疑是否是在复制构造函数上出现问题，但排除了这种可能。难道是CodeBlocks中编译器新版本中的新要求，不而知。先发在此处等读者在评价中指点吧。