【C++深度剖析教程11】C++学习之编写代码实现复数类

今天，我来学习将复数的加减乘除以及比较运算，编写一个复数类，方便计算复数之间的运算。具体用的方法就是之前写过的操作符重载的概念来实现（操作符重载的概念学习）。
那么为了显得清晰，今天写的程序运用模块化的思想：

Complex.h为：

#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

Complex.cpp为：

#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)
{if( this != &c ){a = c.a;b = c.b;}return *this;
}

main.cpp为：

#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;
}