c语言快速复数的比较大小,C语言实现常用的复数运算
这是最近学习的C语言数据结构的中的案例，但是没有给出具体的实现代码。根据自己的学习水平简单编写了一下，倒是能够计算与输出，但是不知道还有没有什么缺陷。
借用到了部分之前博客中展示的代码，以下是今天实现的核心代码：
#include "complex.h"
#include "complexp.h"
CPX_NUM_p_t CPX_compute_sum(CPX_NUM_p_t cpx1,CPX_NUM_p_t cpx2)
{
CPX_NUM_p_t cpx_sum = CDA_NEW(CPX_NUM_t);
cpx_sum->real = cpx1->real + cpx2->real;
cpx_sum->imaginary = cpx1->imaginary + cpx2->imaginary;
return cpx_sum;
}
CPX_NUM_p_t CPX_compute_diff(CPX_NUM_p_t cpx1,CPX_NUM_p_t cpx2)
{
CPX_NUM_p_t cpx_diff = CDA_NEW(CPX_NUM_t);
cpx_diff->real = cpx1->real - cpx2->real;
cpx_diff->imaginary = cpx1->imaginary - cpx2->imaginary;
return cpx_diff;
}
CPX_NUM_p_t CPX_compute_neg(CPX_NUM_p_t cpx)
{
CPX_NUM_p_t result = CDA_NEW(CPX_NUM_t);
result->real = -cpx->real;
result->imaginary = -cpx->imaginary;
return result;
}
CPX_NUM_p_t CPX_compute_prod(CPX_NUM_p_t cpx1,CPX_NUM_p_t cpx2)
{
CPX_NUM_p_t prod = CDA_NEW(CPX_NUM_t);
prod->real = cpx1->real * cpx2->real - cpx1->imaginary * cpx2->imaginary;
prod->imaginary = cpx1->imaginary * cpx2->real + cpx1->real * cpx2->imaginary;
return prod;
}
CPX_NUM_p_t CPX_compute_quot(CPX_NUM_p_t cpx1,CPX_NUM_p_t cpx2)
{
CPX_NUM_p_t quot = CDA_NEW(CPX_NUM_t);
double a,b,c,d;
a = cpx1->real;
b = cpx1->imaginary;
c = cpx2->real;
d = cpx2->imaginary;
quot->real = (a * c + b * d)/(c * c + d * d);
quot->imaginary = (b * c - a * d)/(c * c + d *d);
return quot;
}
主函数以及简单的测试接口代码：
#include "stdio.h"
#include "cda.h"
#include "complex.h"
int main(void)
{
CPX_NUM_t complex_num1;
CPX_NUM_t complex_num2;
CPX_NUM_t result;
complex_num1.real = 123;
complex_num1.imaginary = 234;
complex_num2.real = 345;
complex_num2.imaginary = 456;
result = *CPX_compute_sum(&complex_num1,&complex_num2);
printf("For Sum reuslt is:(%lf,%lfi)\n",result.real,result.imaginary);
result = *CPX_compute_diff(&complex_num1,&complex_num2);
printf("For Diff reuslt is:(%lf,%lfi)\n",result.real,result.imaginary);
result = *CPX_compute_neg(&complex_num1);
printf("For Neg reuslt is:(%lf,%lfi)\n",result.real,result.imaginary);
result = *CPX_compute_prod(&complex_num1,&complex_num2);
printf("For Prod reuslt is:(%lf,%lfi)\n",result.real,result.imaginary);
result = *CPX_compute_quot(&complex_num1,&complex_num2);
printf("For Quot reuslt is:(%lf,%lfi)\n",result.real,result.imaginary);
return 0;
}
编译以及运行结果：
