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。