/* (程序头部注释开始)* 程序的版权和版本声明部分* Copyright (c) 2011, 烟台大学计算机学院学生 * All rights reserved.* 文件名称： * 作 者： 李洪悬 * 完成日期： 2012 年 4 月 11 日* 版 本 号： * 对任务及求解方法的描述部分* 输入描述： * 问题描述：* 程序输出： * 程序头部的注释结束*/
【任务1】实现复数类中的运算符重载定义一个复数类重载运算符+、-、*、/，使之能用于复数的加减乘除。（1）方案一：请用类的成员函数完成运算符的重载；class Complex{public:Complex(){real=0;imag=0;}Complex(double r,double i){real=r;imag=i;}Complex operator+(Complex &c2);Complex operator-(Complex &c2);Complex operator*(Complex &c2);Complex operator/(Complex &c2);void display();private:double real;double imag;};//下面定义成员函数int main(){Complex c1(3,4),c2(5,-10),c3;cout<<"c1=";c1.display();cout<<"c2=";c2.display();c3=c1+c2;cout<<"c1+c2=";c3.display();c3=c1-c2;cout<<"c1-c2=";c3.display();c3=c1*c2;cout<<"c1*c2=";c3.display();c3=c1/c2;cout<<"c1/c2=";c3.display();system("pause");return 0;}（2）方案二：请用类的友元函数，而不是成员函数，完成上面提及的运算符的重载；（3）方案三：在方案二的基础上，扩展+、-、*、/运算符的功能，使之能与double型数据进行运算。设Complex c; double d; c?d和d?c的结果为将d视为实部为d的复数同c运算的结果（其中?为+、-、*、/之一）。另外，定义一目运算符-，-c相当于0-c。#include < iostream > using namespace std; class Complex { public: Complex() { real=0; imag=0; } Complex (double r,double i){real=r;imag=i;} Complex operator - ();//求反函数 friend Complex operator + (Complex &c1 , Complex &c2); friend Complex operator - (Complex &c1 , Complex &c2); friend Complex operator * (Complex &c1 , Complex &c2); friend Complex operator / (Complex &c1 , Complex &c2); friend Complex operator + (double d , Complex &c1);friend Complex operator - (double d , Complex &c1);friend Complex operator * (double d , Complex &c1);friend Complex operator /(double d , Complex &c1);friend Complex operator + (Complex &c1, double d);friend Complex operator - (Complex &c1, double d);friend Complex operator * (Complex &c1, double d);friend Complex operator / (Complex &c1, double d); void display(); private: double real; double imag; }; Complex Complex ::operator - (){ return Complex(0 - this->real,0 - this->imag);//return (0 - *this);更好！}//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; } //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; } //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; } //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; } //浮点型双精度数和复数相加Complex operator + (double d , Complex &c1){return Complex( d+c1.real , c1.imag);}//浮点型双精度数和复数相减Complex operator - (double d , Complex &c1){return Complex( d - c1.real , -c1.imag);}//浮点型双精度数和复数相乘Complex operator * (double d , Complex &c1){Complex c(d ,0);return (c * c1); }//浮点型双精度数和复数相除Complex operator / (double d , Complex &c1){Complex c(d , 0);return ( c / c1);}//浮点型双精度数和复数相加Complex operator + ( Complex &c1, double d ){return Complex(c1.real + d , c1.imag);}//浮点型双精度数和复数相减Complex operator - (Complex &c1, double d){return Complex( c1.real - d , c1.imag);}//浮点型双精度数和复数相乘Complex operator * (Complex &c1, double d){Complex c(d ,0);return (c1 * c); }//浮点型双精度数和复数相除Complex operator / (Complex &c1, double d){Complex c(d , 0);return ( c1 / c);}void Complex::display() { cout << "(" << real << " , " << imag << "i)" << endl; } int main() { Complex c1(3 , 4),c2(5 , -10),c3; double d = 7.5; cout << " d = " << d << endl; cout << " c1 = "; c1.display(); cout << " c2 = "; c2.display(); cout << "对C1和C2求反：" << endl; cout << " - c1 = "; c3 = - c1; c3.display(); cout << " - c2 = "; c3 = - c2; c3.display(); c3 = c1 + c2; cout << " c1 + c2 = "; c3.display(); c3 = c1 - c2; cout << " c1 - c2 = "; c3.display(); c3 = c1 * c2; cout << " c1 * c2 = "; c3.display (); c3 = c1 / c2; cout << " c1 / c2 = "; c3.display (); c3 = d + c1; cout << " d + c1 ="; c3.display(); c3 = d - c1; cout << " d - c1 = "; c3.display(); c3 = d * c1; cout << " d * c1 = "; c3.display (); c3 = d / c1; cout << " d / c1 = "; c3.display (); c3 =c1 + d ; cout << " c1 + d = "; c3.display(); c3 = c1 - d; cout << " c1 - d ="; c3.display(); c3 = c1 * d; cout << " c1 * d ="; c3.display (); c3 = c1 / d; cout << " c1 / d ="; c3.display (); system ( " pause " ); return 0; }
