複數 一級ADT實現

COMPLEX.h

 1 /*  2 typedef struct  3 {  4  float RE; //實部  5  float IM; //虛部  6 }Complex;  7 */  8 typedef struct complex * Complex;  9 10 Complex COMPLEXinit(float, float); 11 float Re(Complex); 12 float Im(Complex); 13 Complex COMPLEXmult(Complex, Complex);

COMPLEX.c

 1 #include "COMPLEX.h"  2  3 struct complex  4 {  5 float RE; //實部  6 float IM; //虛部  7 };  8  9 Complex COMPLEXinit(float RE, float IM) 10 { 11 /* 12  Complex t; 13  t.RE=RE; 14  t.IM=IM; 15  return t; 16 */ 17 18 Complex t=malloc(sizeof *t); 19 t->RE=RE; 20 t->IM=IM; 21 return t; 22 } 23 float Re(Complex z) 24 { 25 return z->RE; 26 } 27 float Im(Complex z) 28 { 29 return z->IM; 30 } 31 Complex COMPLEXmult(Complex a, Complex b) 32 { 33 /* 34  Complex t; 35  t.RE=a.RE*b.RE-a.IM*b.IM; 36  t.IM=a.RE*b.IM+a.IM*b.RE; 37 38  //a實部乘b實部-a虛部乘b虛部 39  //a實部乘b虛部+a虛部乘b實部 40  return t;*/ 41 42 return COMPLEXinit(Re(a)*Re(b)-Im(a)*Im(b), 43 Re(a)*Im(b)+Im(a)*Re(b)); 44 }

main.c

 1 #include <stdio.h>  2 #include <math.h>  3 #include "COMPLEX.h"  4  5 #define PI 3.141592625  6  7 int main(void)  8 {  9 int N; 10 printf("輸入一個參數:"); 11 scanf("%d", &N); 12  getchar(); 13 14  Complex t, x; 15 printf("%dth complex roots of unity\n", N); 16 for(int i=0; i<N; i++) 17  { 18 float r=2.0*PI*i/N; 19 //1=e^(2n*pi*i) ? 20 21 22 t=COMPLEXinit(cos(r), sin(r)); 23 24 printf("%2d %6.3f %6.3f ", i, Re(t), Im(t)); 25 x=t; 26 for(int j=0; j<N-1; j++) 27 x=COMPLEXmult(t, x); 28 29 printf("%6.3f %6.3f\n", Re(x), Im(x)); 30  } 31 32 return 0; 33 }