复数 一级ADT实现

COMPLEX.h

 /*
 typedef struct
 {
     float RE; //实部
     float IM; //虚部
 }Complex;
 */
 typedef struct complex * Complex;

 Complex COMPLEXinit(float, float);
 float Re(Complex);
 float Im(Complex);
 Complex COMPLEXmult(Complex, Complex);

COMPLEX.c

 #include "COMPLEX.h"

 struct complex
 {
     float RE; //实部
     float IM; //虚部
 };

 Complex COMPLEXinit(float RE, float IM)
 {
     /*
     Complex t;
     t.RE=RE;
     t.IM=IM;
     return t;
     */

     Complex t=malloc(sizeof *t);
     t->RE=RE;
     t->IM=IM;
     return t;
 }
 float Re(Complex z)
 {
     return z->RE;
 }
 float Im(Complex z)
 {
     return z->IM;
 }
 Complex COMPLEXmult(Complex a, Complex b)
 {
     /*
     Complex t;
     t.RE=a.RE*b.RE-a.IM*b.IM;
     t.IM=a.RE*b.IM+a.IM*b.RE;

     //a实部乘b实部-a虚部乘b虚部
     //a实部乘b虚部+a虚部乘b实部
     return t;*/

     return COMPLEXinit(Re(a)*Re(b)-Im(a)*Im(b),
                        Re(a)*Im(b)+Im(a)*Re(b));
 }

main.c

 #include <stdio.h>
 #include <math.h>
 #include "COMPLEX.h"

 #define PI 3.141592625

 int main(void)
 {
     int N;
     printf("输入一个参数:");
     scanf("%d", &N);
     getchar();

     Complex t, x;
     printf("%dth complex roots of unity\n", N);
     for(int i=; i<N; i++)
     {
         float r=2.0*PI*i/N;
         //1=e^(2n*pi*i)  ?

         t=COMPLEXinit(cos(r), sin(r));

         printf("%2d %6.3f %6.3f ", i, Re(t), Im(t));
         x=t;
         for(int j=; j<N-; j++)
             x=COMPLEXmult(t, x);

         printf("%6.3f %6.3f\n", Re(x), Im(x));
     }

     return ;
 }