Q：下面的复数解决方案是否可行？

class Complex{ public: int a; int b;};int main(){ Complex c1={1,2}; Complex c2={3,4}; Complex c3=c1+c2; return 0;}

该段代码想要实现的是将两个复数类进行相加得出第三个类
代码实现的运行结果

由上面的结果图可以得知，出现的错误是无法匹配+号操作符的操作，同时出现 的潜在问题是a与b是public成员，在实际的操作中应将a与b设置为private成员
改正的代码示例

#include <iostream>using namespace std;class Complex { int a; int b;public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } friend Complex Add(const Complex& p1, const Complex& p2);};Complex Add(const Complex& p1, const Complex& p2){ Complex ret; ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret;}int main(){ Complex c1(1, 2); Complex c2(3, 4); Complex c3 = Add(c1, c2); // c1 + c2 cout<<"c3.a ="<<c3.getA()<<endl; cout<<"c3.b ="<<c3.getB()<<endl; return 0;}

该代码运行了友元函数friend，同时定义了全局函数Add,将a与b设置为私有成员
运行结果

出现的疑问：Add函数可以解决Complex对象相加的问题，但是Complex是现实世界中确实存在的复数，并且复数在数学中的地位和普通的实数相同---为什么不能让+操作符也支持复数相加？
操作符重载
1.C++中的重载能够扩展操作符的功能
2.操作符的重载以函数的方式进行
本质--用特殊形式的函数扩展操作符的功能
通过operator关键字可以定义特殊的函数
operator的本质是通过函数重载操作符
语法

操作符重载示例

#nclude <isotream>using namespace std;class Complex { int a; int b;public: Complex(int a = 0, int b = 0) { this->a = a; this->b = b; } int getA() { return a; } int getB() { return b; } friend Complex operator + (const Complex& p1, const Complex& p2);};Complex operator + (const Complex& p1, const Complex& p2){ Complex ret; ret.a = p1.a + p2.a; ret.b = p1.b + p2.b; return ret;}int main(){ Complex c1(1, 2); Complex c2(3, 4); Complex c3 = c1 + c2; // operator + (c1, c2) cout<<"c3.a ="<<c3.getA()<<endl; cout<<"c3.b ="<<c3.getB()<<endl; return 0;}

运行结果如图所示

可以将操作符重载定义为类的成员函数
1.比全局操作符重载函数少一个参数
2.不需要依赖友元就可以完成操作符重载
3.编译器优先在成员函数中寻找操作符重载函数

小结
1.操作符重载是C++的强大特性之一
2.操作符重载的本质是通过函数扩展操作符的功能
3.operator关键字是实现操作符重载的关键
4.操作符重载遵循相同的函数重载规则
5.全局函数和成员函数都可以实现对操作符的重载

复数类应该具有的操作

利用操作符重载
1.统一复数与实数的运算方式
2.统一复数与实数的比较方式

double getModulus(); Complex operator + (const Complex& c); Complex operator - (const Complex& c); Complex operator * (const Complex& c); Complex operator / (const Complex& c); bool operator == (const Complex& c); bool operator != (const Complex& c); Complex& operator = (const Complex& c);

复数类的实现
Complex.cpp

#include "Complex.h"#include "math.h"Complex::Complex(double a, double b){ this->a = a; this->b = b;}double Complex::getA(){ return a;}double Complex::getB(){ return b;}double Complex::getModulus(){ return sqrt(a * a + b * b);}Complex Complex::operator + (const Complex& c){ double na = a + c.a; double nb = b + c.b; Complex ret(na, nb); return ret;}Complex Complex::operator - (const Complex& c){ double na = a - c.a; double nb = b - c.b; Complex ret(na, nb); return ret;}Complex Complex::operator * (const Complex& c){ double na = a * c.a - b * c.b; double nb = a * c.b + b * c.a; Complex ret(na, nb); return ret;}Complex Complex::operator / (const Complex& c){ double cm = c.a * c.a + c.b * c.b; double na = (a * c.a + b * c.b) / cm; double nb = (b * c.a - a * c.b) / cm; Complex ret(na, nb); return ret;}bool Complex::operator == (const Complex& c){ return (a == c.a) && (b == c.b);}bool Complex::operator != (const Complex& c){ return !(*this == c);}Complex& Complex::operator = (const Complex& c)//返回值是个引用{ if( this != &c ) { a = c.a; b = c.b; } return *this;}

Complex.h

#ifndef _COMPLEX_H_#define _COMPLEX_H_class Complex{ double a; double b;public: Complex(double a = 0, double b = 0); double getA(); double getB(); double getModulus(); Complex operator + (const Complex& c); Complex operator - (const Complex& c); Complex operator * (const Complex& c); Complex operator / (const Complex& c); bool operator == (const Complex& c); bool operator != (const Complex& c); Complex& operator = (const Complex& c);};#endif

test.cpp

#include <stdio.h>#include "Complex.h"int main(){ Complex c1(1, 2); Complex c2(3, 6); Complex c3 = c2 - c1; Complex c4 = c1 * c3; Complex c5 = c2 / c1; printf("c3.a = %f, c3.b = %f\n", c3.getA(), c3.getB()); printf("c4.a = %f, c4.b = %f\n", c4.getA(), c4.getB()); printf("c5.a = %f, c5.b = %f\n", c5.getA(), c5.getB()); Complex c6(2, 4); printf("c3 == c6 : %d\n", c3 == c6); printf("c3 != c4 : %d\n", c3 != c4); (c3 = c2) = c1; printf("c1.a = %f, c1.b = %f\n", c1.getA(), c1.getB()); printf("c2.a = %f, c2.b = %f\n", c2.getA(), c2.getB()); printf("c3.a = %f, c3.b = %f\n", c3.getA(), c3.getB()); return 0;}

输出结果

注意事项
1.C++规定赋值操作符(=)只能重载为成员函数
2.操作符重载不能改变原操作符的优先级
3.操作符重载不能改变操作的个数
4.操作符重载不应该改变操作符的原有语义

小结
1.复数的概念可以通过自定义类实现
2.复数中的运算操作可以通过操作符重载来实现
3.赋值操作符只能通过成员函数实现
4.操作符重载的本质为函数定义