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)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;
}
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。
