《C++语言基础》实践参考——复数类中的运算符重载(续)

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【项目1-复数类中的运算符重载(续)】
复数类中的运算符重载基础上
(1)再定义一目运算符 -,-c相当于0-c。
(2)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]

#include <iostream>
using namespace std;
class Complex
{
public:
    Complex()
    {
        real=0;
        imag=0;
    }
    Complex(double r,double i)
    {
        real=r;
        imag=i;
    }
    Complex operator-();
    //实现输入、输出的运算符重载
    friend ostream& operator << (ostream& output, const 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, const 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-*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 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)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;
    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;
}




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