C Primer Plus（第六版）16.18 编程练习 第3题

#include <stdio.h>

#include <math.h>

#define PI 3.14159265358979

struct j{

double len;

double jd;

};

struct z{

double x;

double y;

};

struct z jtoz(struct j pj);

int main(void)

{

double hd;

struct j pj = {

sqrt(2.0),//根号2,45度的边长是1

45.0,

};

struct z zj;

zj=jtoz(pj);

printf("x=%lf y=%lf\n",zj.x,zj.y);

}

struct z jtoz(struct j pj)

{

struct z zj;

double hd;

hd=pj.jd*PI/180.0;

zj.x = pj.len * cos(hd);

zj.y = pj.len * sin(hd);

return zj;

}