重载乘法运算符

Question

我正在为我的 c++ class 做作业。我们不得不重载几个运算符，例如 +、-、!=、= 等。好吧，除了乘法，我已经弄清楚了所有这些运算符。我尝试过的所有内容都会溢出或无法编译。不确定我需要什么。

这是包含我的重载的头文件。

#ifndef COMPLEXNUMBER_H
#define COMPLEXNUMBER_H
#include<iostream>
using namespace std;

class ComplexNumber{
public:

    double real, imaginary;
    ComplexNumber(){
        real = 0;
        imaginary = 0;
    }
    ComplexNumber(double a, double b){
        real = a;
        imaginary = b;
    }
    ComplexNumber(double a){

        real = a;
        imaginary = 0;

    }

    ComplexNumber & operator= (const ComplexNumber & rhs){

        if(this == &rhs){

            return *this;

        }
        else{

            real = rhs.imaginary;
            imaginary = rhs.imaginary;
        }

        return *this;
    }

    ComplexNumber & operator+= (const ComplexNumber &rhs){

        real += rhs.real;
        imaginary += rhs.imaginary;
        return *this;

    }

    ComplexNumber & operator-= (const ComplexNumber &rhs){

        real -= rhs.real;
        imaginary -= rhs.imaginary;
        return *this;

    }

    const ComplexNumber operator+ (const ComplexNumber &rhs){

        ComplexNumber result = *this;
        result += rhs;
        return result;

    }

    const ComplexNumber operator- (const ComplexNumber &rhs){

        ComplexNumber result = *this;
        result -= rhs;
        return result;

    }

    const ComplexNumber operator* (const ComplexNumber &rhs){

        return *this * rhs;

    }

    friend ostream & operator<< (ostream &out, const ComplexNumber &c){

        out << "(" << c.real << (c.imaginary<0?" - ":" + ") << abs(c.imaginary) << " i)";
        return out;

    }

    friend istream & operator>> (istream & in, ComplexNumber &c){

        in >> c.real >> c.imaginary;
        return in;

    }

    operator double(){

        return real;

    }

    bool operator== (const ComplexNumber & rhs) const {

        bool result = (this->real == rhs.real) && (this->imaginary == rhs.imaginary);
        return result;

    }

    bool operator!= (const ComplexNumber &rhs) const{

        return !(*this == rhs);

    }
};

#endif

我知道乘法运算符还有很长的路要走，但这正是我目前所拥有的。在这里它是独立的。任何想法将不胜感激！！

const ComplexNumber operator* (const ComplexNumber &rhs){

            return *this * rhs;

        }

Answer 1

由于您对它的调用方式，它给您带来了溢出。通过您的调用，您将一个复数与一个复数相乘，它只是不断调用同一个运算符而不做任何事情。您可以尝试使用一些基本数学并推导出复数乘法的公式。具体来说，假设我们有两个复数 Z1 和 Z2。令 Z1 = a + bi，其中 a 为实部，b 为虚部，Z2 = c + di，其中 c 为实部，d 为虚部。我们有 Z1 * Z2 = (a + bi)(c + di) = ( ac + adi + cbi - bd )。现在，我们将实部和虚部分开，这里的实部是没有 i 的所有内容，所以 ac - bd，虚部是 ad + cb。现在，在您的 class 成员的条款中使用它，您将得到如下内容：

const ComplexNumber operator* (const ComplexNumber &rhs)
{
    ComplexNumber result;
    result.real = real * rhs.real - imaginary * rhs.imaginary;
    result.imaginary = real * rhs.imaginary + imaginary * rhs.real;
    return result;
}

重载乘法运算符

Overloading Multiplication Operator

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operator-overloading

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