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C语言实现直角坐标转换为极坐标的方法

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本文实例讲述了C语言实现直角坐标转换为极坐标的方法。分享给大家供大家参考,具体如下: #includestdio.h#includemath.hstruct complex_s{ double x,y;};double real_part(struct complex_s z){ return z.x;}doubl

本文实例讲述了C语言实现直角坐标转换为极坐标的方法。分享给大家供大家参考,具体如下:

#include<stdio.h>
#include<math.h>
struct complex_s{
    double x,y;
};
double real_part(struct complex_s z){
    return z.x;
}
double img_part(struct complex_s z){
    return z.y;
}
double magnitude(struct complex_s z){
    return sqrt(z.x*z.x + z.y*z.y);
}
double angle(struct complex_s z){
    return atan2(z.y, z.x);
}
struct complex_s make_from_real_img(double x, double y){
    struct complex_s z;
    z.x = x;
    z.y = y;
    return z;
}
struct complex_s make_from_mag_ang(double r, double A){
    struct complex_s z;
    z.x = r * cos(A);
    z.y = r * sin(A);
    return z;
}
struct complex_s add_complex(struct complex_s z1,struct complex_s z2){
    return make_from_real_img(real_part(z1)+real_part(z2),
        img_part(z1) + img_part(z2));
}
struct complex_s sub_complex(struct complex_s z1,struct complex_s z2){
    return make_from_real_img(real_part(z1)-real_part(z2),
        img_part(z1) - img_part(z2));
}
struct complex_s mul_complex(struct complex_s z1,struct complex_s z2){
    return make_from_mag_ang(real_part(z1)*real_part(z2),
        img_part(z1) + img_part(z2));
}
struct complex_s div_complex(struct complex_s z1,struct complex_s z2){
    return make_from_mag_ang(real_part(z1)/real_part(z2),
        img_part(z1) + img_part(z2));
}
int main(void){
    struct complex_s z1 = {3.0,4.0};
    struct complex_s z2= {2.0,5.0};
    struct complex_s x;
    x = add_complex(z1,z2);
    printf("x={%f,%f}",x.x,x.y);
    return 0;
}

结果:

[root@localhost struct]# ./jizuobiao.out
x={5.000000,9.000000}

希望本文所述对大家C语言程序设计有所帮助。

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