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 ;

 }