接:第8周-任务1-方案2-复数类中运算符重载(非成员函数实现)
本文在方案2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,再定义一目运算符 -,-c相当于0-c。
【讲解视频】
【参考解答】
在下面的解答中,我将所有二目运算符的重载定义为友元函数,一目运算符重载为成员函数,这是惯用的做法。
#include <iostream> using namespace std; class Complex { public: Complex(){real=0;imag=0;} Complex(double r,double i){real=r;imag=i;} Complex operator-(); 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); void display(); private: double real; double imag; }; 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; } void Complex::display() { cout<<"("<<real<<","<<imag<<"i)"<<endl; } int main() { Complex c1(3,4),c2(5,-10),c3; double d=11; cout<<"c1="; c1.display(); cout<<"c2="; c2.display(); cout<<"d="<<d<<endl; cout<<"-c1=";(-c1).display(); c3=c1+c2; cout<<"c1+c2="; c3.display(); cout<<"c1+d="; (c1+d).display(); cout<<"d+c1="; (d+c1).display(); c3=c1-c2; cout<<"c1-c2="; c3.display(); cout<<"c1-d="; (c1-d).display(); cout<<"d-c1="; (d-c1).display(); c3=c1*c2; cout<<"c1*c2="; c3.display(); cout<<"c1*d="; (c1*d).display(); cout<<"d*c1="; (d*c1).display(); c3=c1/c2; cout<<"c1/c2="; c3.display(); cout<<"c1/d="; (c1/d).display(); cout<<"d/c1="; (d/c1).display(); system("pause"); return 0; }
【关于参数类型及限定词选择的一点讨论】
我在空间中,看到了leihengxin发的一个贴子:《链接》
他给的程序是:
#include<iostream> using namespace std; class Complex { public: Complex(){real=0;imag=0;} Complex(double r,double i){real=r;imag=i;} friend Complex operator+(Complex &c1,Complex &c2); friend Complex operator + (Complex &c, double &d); friend Complex operator + (double &d, Complex &c); friend void display(Complex &c2); private: double real; double imag; }; //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+(Complex &c,double &d) { return Complex(c.real+d, c.imag); } Complex operator+(double &d, Complex &c) { return Complex(c.real+d, c.imag); } void display(Complex &c) { cout<<"("<<c.real<<","<<c.imag<<"i)"<<endl; } int main() { Complex c1(3,4),c2(5,-10),c3,c4; double dd=4.32; c3=c1+3.14; cout<<"c1+3.14="; display(c3); c4=3.14+c1; cout<<"3.14+c1="; display(c4); system("pause"); return 0; }
在这个程序中,第X,X行将参数d声明为引用:double &d。这是一个不错的考虑,但也带来了错误,原因在于编译器不能将c1+3.14中3.14这个double型常量与double型引用变量&d匹配起来。至于为什么,请感兴趣的同学能够找资料做些进一步的解释,写在本文的评论中。
如何解决?
有网友给出了一种解决方案:
(1)函数参数不要用引用
friend Complex operator + (Complex &c1, doubled); friend Complex operator + (double d,Complex &c1);
这种方法可行。但d用成引用类型并无不妥。保持d为引用,还有两种改法:
(2)在main()函数中,声明一个double型变量,如double dd=3.14,然后再执行c3=c1+dd。这能使显然这种方法给重载的运算符的使用套上了枷锁(只能加变量不能加常量,要让复数加一个常量,必然先要赋值给一个变量),并不可取。
(3)将运算符重载函数的参数加上const限定词,为double型的引用const double &d
friend Complex operator + (Complex &c,constdouble &d); friend Complex operator+ (const double&d, Complex &c);
这样,调用c1+3.14时,编译器顺利地将3.14解释为double型常量。使用这种方案,支持复数加double型变量,而且,当d为变量,在函数中是不允许修改d值,这与“执行加法”的要求是一致的,我们不希望完成操作了,加数也发生了变化。甚至在这里,将参数c也声明为const更好,即
friend Complex operator + (const Complex &c, const double &d); friend Complex operator+ (const double&d, const Complex &c);
所以,在以上的三种方案中,倾向于用第三种方案。
再讨论一个钻牛角尖的问题,如果采用第三种方案,既能加常量,也能加变量,而且当加变量时,变量值在函数体内是可变的(当然这种设计并不可取),该如何去做。我们可以继续重载operate+函数,一共需要4个版本的定义了:
friend Complex operator + (const Complex &c, constdouble &d); friend Complex operator+ (const double&d, const Complex &c); friend Complex operator + (Complex &c, double &d); friend Complex operator+ (double &d, Complex &c);
加与不加const是有区别的,编译器认可这是不同的函数,在参数是非const变量时,爱改就改吧。