深圳哪家网站建设的公司好山东seo推广

在本文中，我将向您展示一种使用C编程语言实现复数的方法。 请注意，使用此处提供的代码没有任何保证。

复数是一个数字，例如z，表示z = realPart + i * imaginaryPart，其中i是虚数单位，有时用j表示。 另外，i * i = -1，这在查找两个复数的乘积/除法时很重要。

自然，我们可以将复数实现为结构：

/* Definition of a Complex number: */
typedef struct complex {double realPart;double imaginaryPart;
} Complex;

我们可以按如下方式将接口与实现分开。

首先，我们有标题

Complex.h ，它定义了Complex结构，并包含处理一个或多个Complex数字的函数的原型：

/* Complex.h */ 
#ifndef COMPLEX_H
#define COMPLEX_H 
/* Definition of a Complex number: */
typedef struct complex {double realPart;double imaginaryPart;
} Complex; 
/* Prototypes for functions to manipulate Complex number(s) (these functions are implemented in file Complex.c): */
Complex addition( const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 );
Complex subtraction( const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 );
Complex multiplication( const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 );
Complex division(  const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 );
void print( const Complex *pointerToComplexNumber ); 
#endif

文件Complex.c实现了以上功能：

/* Complex.c */ 
#include <stdio.h>
#include "Complex.h" 
/* Adds two Complex numbers, returning a new one, without modifying the ones received: */
Complex addition( const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 )
{Complex complexNumber3; complexNumber3.realPart = pointerToComplexNumber1->realPart + pointerToComplexNumber2->realPart;complexNumber3.imaginaryPart = pointerToComplexNumber1->imaginaryPart + pointerToComplexNumber2->imaginaryPart; return complexNumber3;
} 
/* Subtracts two Complex numbers, returning a new one, without modifying the ones received: */
Complex subtraction( const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 )
{Complex complexNumber3; complexNumber3.realPart = pointerToComplexNumber1->realPart - pointerToComplexNumber2->realPart;complexNumber3.imaginaryPart = pointerToComplexNumber1->imaginaryPart - pointerToComplexNumber2->imaginaryPart; return complexNumber3;
} 
/* Multiplies two Complex numbers, returning a new one, without modifying the ones received: */
Complex multiplication( const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 )
{Complex complexNumber3; complexNumber3.realPart = pointerToComplexNumber1->realPart * pointerToComplexNumber2->realPart -pointerToComplexNumber1->imaginaryPart * pointerToComplexNumber2->imaginaryPart; complexNumber3.imaginaryPart = pointerToComplexNumber1->realPart * pointerToComplexNumber2->imaginaryPart +pointerToComplexNumber1->imaginaryPart * pointerToComplexNumber2->realPart; return complexNumber3;
} 
/* Divides two Complex numbers, returning a new one, without modifying the ones received: */
Complex division(  const Complex *pointerToComplexNumber1, const Complex *pointerToComplexNumber2 )
{Complex complexNumber3;double commonDenominator = pointerToComplexNumber2->realPart * pointerToComplexNumber2->realPart +pointerToComplexNumber2->imaginaryPart * pointerToComplexNumber2->imaginaryPart; complexNumber3.realPart = ( pointerToComplexNumber1->realPart * pointerToComplexNumber2->realPart +pointerToComplexNumber1->imaginaryPart * pointerToComplexNumber2->imaginaryPart ) / commonDenominator; complexNumber3.imaginaryPart = ( pointerToComplexNumber1->imaginaryPart * pointerToComplexNumber2->realPart -pointerToComplexNumber1->realPart * pointerToComplexNumber2->imaginaryPart ) / commonDenominator; return complexNumber3;
} 
/* Prints a Complex number in the form ( realPart ) + i( imaginaryPart ), without modifying it: */
void print( const Complex *pointerToComplexNumber )
{printf( "( %.4f ) + i( %.4f )", pointerToComplexNumber->realPart, pointerToComplexNumber->imaginaryPart );
}

最后，我们有一个名为ComplexTest.c的文件，该文件在主函数中测试文件Complex.c和Complex.h ，这是我们C程序的起点：

/* ComplexTest.c */ 
#include <stdio.h>
#include "Complex.h" 
int main( void )
{Complex number1;Complex number2;Complex sumOfNumber1WithNumber2;Complex subtractionOfNumber2FromNumber1;Complex multiplicationOfNumber1WithNumber2;Complex divisionOfNumber1PerNumber2; number1.realPart = 1.2343;number1.imaginaryPart = 4.7621; number2.realPart = -3.213;number2.imaginaryPart = -9.8; sumOfNumber1WithNumber2 = addition( &number1, &number2 );subtractionOfNumber2FromNumber1 = subtraction( &number1, &number2 );multiplicationOfNumber1WithNumber2 = multiplication( &number1, &number2 );divisionOfNumber1PerNumber2 = division( &number1, &number2 ); printf( "Number1 = " );print( &number1 );printf( "\nNumber2 = " );print( &number2 );printf( "\nNumber1 + Number2 = " );print( &sumOfNumber1WithNumber2 );printf( "\nNumber1 - Number2 = " );print( &subtractionOfNumber2FromNumber1 );printf( "\nNumber1 * Number2 = " );print( &multiplicationOfNumber1WithNumber2 );printf( "\nNumber1 / Number2 = " );print( &divisionOfNumber1PerNumber2 );putchar( '\n' ); return 0;
}

要使用Visual Studio 2013编译该程序，请从命令行打开VS2013的“开发人员命令提示符”，转到上述三个文件所在的目录，然后输入命令cl ComplexTest.c Complex.c。

然后，使用以下命令运行程序

综合测试

From: https://bytes.com/topic/c/insights/956447-implementation-complex-numbers-c