第31课 完善的复数类

Posted 2021-01-03 wanmeishenghuo

复数类：

程序如下：

 1 #ifndef _COMPLEX_H_
 2 #define _COMPLEX_H_
 3 
 4 class Complex
 5 {
 6     double a;
 7     double b;
 8 public:
 9     Complex(double a = 0, double b = 0);
10     double getA();
11     double getB();
12     double getModulus();
13     
14     Complex operator + (const Complex& c);
15     Complex operator - (const Complex& c);
16     Complex operator * (const Complex& c);
17     Complex operator / (const Complex& c);
18     
19     bool operator == (const Complex& c);
20     bool operator != (const Complex& c);
21     
22     Complex& operator = (const Complex& c);
23 };
24 
25 #endif

 1 #include "Complex.h"
 2 #include "math.h"
 3 
 4 Complex::Complex(double a, double b)
 5 {
 6     this->a = a;
 7     this->b = b;
 8 }
 9 
10 double Complex::getA()
11 {
12     return a;
13 }
14 
15 double Complex::getB()
16 {
17     return b;
18 }
19 
20 double Complex::getModulus()
21 {
22     return sqrt(a * a + b * b);
23 }
24 
25 Complex Complex::operator + (const Complex& c)
26 {
27     double na = a + c.a;
28     double nb = b + c.b;
29     Complex ret(na, nb);
30     
31     return ret;
32 }
33 
34 Complex Complex::operator - (const Complex& c)
35 {
36     double na = a - c.a;
37     double nb = b - c.b;
38     Complex ret(na, nb);
39     
40     return ret;
41 }
42 
43 Complex Complex::operator * (const Complex& c)
44 {
45     double na = a * c.a - b * c.b;
46     double nb = a * c.b + b * c.a;
47     Complex ret(na, nb);
48     
49     return ret;
50 }
51 
52 Complex Complex::operator / (const Complex& c)
53 {
54     double cm = c.a * c.a + c.b * c.b;
55     double na = (a * c.a + b * c.b) / cm;
56     double nb = (b * c.a - a * c.b) / cm;
57     Complex ret(na, nb);
58     
59     return ret;
60 }
61     
62 bool Complex::operator == (const Complex& c)
63 {
64     return (a == c.a) && (b == c.b);
65 }
66 
67 bool Complex::operator != (const Complex& c)
68 {
69     return !(*this == c);
70 }
71     
72 Complex& Complex::operator = (const Complex& c)
73 {
74     if( this != &c )
75     {
76         a = c.a;
77         b = c.b;
78     }
79     
80     return *this;
81 }

 1 #include <stdio.h>
 2 #include "Complex.h"
 3 
 4 int main()
 5 {
 6     Complex c1(1, 2);
 7     Complex c2(3, 6);
 8     Complex c3 = c2 - c1;
 9     Complex c4 = c1 * c3;
10     Complex c5 = c2 / c1;
11     
12     printf("c3.a = %f, c3.b = %f
", c3.getA(), c3.getB());
13     printf("c4.a = %f, c4.b = %f
", c4.getA(), c4.getB());
14     printf("c5.a = %f, c5.b = %f
", c5.getA(), c5.getB());
15     
16     Complex c6(2, 4);
17     
18     printf("c3 == c6 : %d
", c3 == c6);
19     printf("c3 != c4 : %d
", c3 != c4);
20     
21     (c3 = c2) = c1;
22     
23     printf("c1.a = %f, c1.b = %f
", c1.getA(), c1.getB());
24     printf("c2.a = %f, c2.b = %f
", c2.getA(), c2.getB());
25     printf("c3.a = %f, c3.b = %f
", c3.getA(), c3.getB());
26     
27     return 0;
28 }

为了能实现连续的赋值操作，第72行我们返回的是复数的引用。

主函数的第21行先把c2赋值给c3，然后返回c3，最后将c1赋值给c3，返回c3。

 运行结果如下：

注意事项：

 

 小结：