[問題] 關於使用SWITCH與檔案讀寫的問題

[wodada (wodada)]

本人目前正在寫一個關於阻尼震盪的問題，

老師要求我們一定要用到switch指令和file I/O跑出來

本人已經將所有程式都寫好了，執行也沒有錯誤，可是在ＥＸＣＥＬ上

始終看不到數值所在，我試了很久一直不知道原因為何？

想請各位高手能幫我解答一下

程式碼如下：

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

//宣告副程式
double dis1(double u0, double v0, double wn, double zeta0, double ti);
double vel1(double u0, double v0, double wn, double zeta0, double ti);
double dis2(double u0, double v0, double wn, double zeta0, double ti);
double vel2(double u0, double v0, double wn, double zeta0, double ti);
double dis3(double u0, double v0, double wn, double zeta0, double ti);
double vel3(double u0, double v0, double wn, double zeta0, double ti);
double dis4(double u0, double v0, double wn, double zeta0, double ti);
double vel4(double u0, double v0, double wn, double zeta0, double ti);

//定義圓周率
//定義四個圖的zeta數值
#define EPS 　　0.001
#define PI      3.1415926
#define zeta1 　0.02
#define zeta2 　0.05
#define zeta3 　0.1
#define zeta4 　0.2


int main(int argc, char* argv[])
{
   FILE *f1;
   int choice= 0;
   double mm,              //mass
   kk,                     //stiffness
   wn,                     //natural freqency
   cc,                     //damping
   u0,                     //initial displacement
   v0,                     //initial velocity
   T,                      //natural period
  time,           　　　　　//time
  dis_ti,         //displacement at time ti
  vel_ti,         //velocity at time ti
  dis_ti2,
  vel_ti2,
  dis_ti3,
  vel_ti3,
  dis_ti4,
  vel_ti4;

　mm= 1;          //題目給質量,彈簧係數和阻尼
  cc= 0.1;
  kk= 1;
  wn= sqrt(kk/mm);
  T= 2*PI/wn;
  u0= 10;         //初始位置和初速度
  v0= 0;

f1= fopen("show_time.csv", "w+");
if(f1==NULL)
{
  printf("Cannot open show_time.csv, is your EXCEL open?\n");
  exit(0);
}

fprintf(f1, "time, distance, velocity, distance2, velocity2, distance3, velocity3, distance4, velocity4,\n");


switch(choice)
{
  for(time= 0.0;time<= 100.0;time+= 1){
  case 1:
     dis_ti= dis1(u0, v0, wn, zeta1, time);
     vel_ti= vel1(u0, v0, wn, zeta1, time);
     fprintf(f1, "%f, %f, %f,\n", time, dis_ti, vel_ti);
     break;

  case 2:
     dis_ti2= dis2(u0, v0, wn, zeta2, time);
     vel_ti2= vel2(u0, v0, wn, zeta2, time);
　   fprintf(f1, "%f, %f, %f,\n", time, dis_ti2, vel_ti2);
     break;

  case 3:
     dis_ti3= dis3(u0, v0, wn, zeta3, time);
　   vel_ti3= vel3(u0, v0, wn, zeta3, time);
 　  fprintf(f1, "%f, %f, %f,\n", time, dis_ti3, vel_ti3);
     break;

  case 4:
　   dis_ti4= dis4(u0, v0, wn, zeta4, time);
  　 vel_ti4= vel4(u0, v0, wn, zeta4, time);
     fprintf(f1, "%f, %f, %f,\n", time, dis_ti4, vel_ti4);
     break;
　  }
　}

　　　　　　　　fclose(f1);
                return 0;
}

//用副程式計算方程式各個係數求出位置
double dis1(double u0, double v0, double wn, double zeta0, double ti)
{
  double wd, A, B, C, D, E;
  wd= wn*sqrt(1-pow(zeta1,2));
  A= exp(-zeta1*wn*ti);
  B= u0*cos(wd*ti);
  C= (v0+wn*zeta1*u0)/wd;
  D= C*sin(wd*ti);
  E= A*(B+D);
 return E;
}

//用副程式計算速度
double vel1(double u0, double v0, double wn, double zeta0, double ti)
{
  double dis_t1, dis_t2, vel_ti;
  dis_t1= dis1(u0, v0, wn, zeta1, ti);
  dis_t2= dis1(u0, v0, wn, zeta1, ti+EPS);
  vel_ti= (dis_t1-dis_t2)/EPS;
  return(vel_ti);
}

double dis2(double u0, double v0, double wn, double zeta0, double ti)
{
  double wd2, A2, B2, C2, D2, E2;
  wd2= wn*sqrt(1-pow(zeta2,2));
  A2= exp(-zeta2*wn*ti);
  B2= u0*cos(wd2*ti);
  C2= (v0+wn*zeta2*u0)/wd2;
  D2= C2*sin(wd2*ti);
  E2= A2*(B2+D2);
  return E2;
}

double vel2(double u0, double v0, double wn, double zeta0, double ti)
{
  double dis_t12, dis_t22, vel_ti2;
  dis_t12= dis3(u0, v0, wn, zeta2, ti);
  dis_t22= dis3(u0, v0, wn, zeta2, ti+EPS);
　vel_ti2= (dis_t12-dis_t22)/EPS;
  return(vel_ti2);
}

double dis3(double u0, double v0, double wn, double zeta0, double ti)
{
  double wd3, A3, B3, C3, D3, E3;
  wd3= wn*sqrt(1-pow(zeta3,2));
  A3= exp(-zeta3*wn*ti);
  B3= u0*cos(wd3*ti);
  C3= (v0+wn*zeta3*u0)/wd3;
  D3= C3*sin(wd3*ti);
  E3= A3*(B3+D3);
  return E3;
}

double vel3(double u0, double v0, double wn, double zeta0, double ti)
{
  double dis_t13, dis_t23, vel_ti3;
  dis_t13= dis3(u0, v0, wn, zeta3, ti);
  dis_t23= dis3(u0, v0, wn, zeta3, ti+EPS);
  vel_ti3= (dis_t13-dis_t23)/EPS;
  return(vel_ti3);
}

double dis4(double u0, double v0, double wn, double zeta0, double ti)
{
  double wd4, A4, B4, C4, D4, E4;
  wd4= wn*sqrt(1-pow(zeta4,2));
  A4= exp(-zeta4*wn*ti);
  B4= u0*cos(wd4*ti);
  C4= (v0+wn*zeta4*u0)/wd4;
  D4= C4*sin(wd4*ti);
  E4= A4*(B4+D4);
  return E4;
}

double vel4(double u0, double v0, double wn, double zeta0, double ti)
{
  double dis_t14, dis_t24, vel_ti4;
  dis_t14= dis3(u0, v0, wn, zeta4, ti);
  dis_t24= dis3(u0, v0, wn, zeta4, ti+EPS);
  vel_ti4= (dis_t14-dis_t24)/EPS;
  return(vel_ti4);
}

程式雖然看起來很多，可是卻很瑣碎，我目前遇到的困難是無法用SWITCH

配合fprintf指令把四個選項的數值輸出到EXCEL上，而輸出到EXCEL上的規格如下：

第一排  第二排  第三排  第四排   第五排.......
                                                    以此類推.....
time    dis_ti  vel_ti  dis_ti2  vel_ti2.......

想請問我該怎麼做才會有此結果產生？







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◆ From: 114.39.17.138
→ wodada:最主要是switch和fprintf部分...
→ loveme00835:你可以用置底文網址貼程式碼, 比較好看
→ wodada:請問一下，是將程式碼丟進該網站後跑出來再複製嗎？
推 VictorTom:貼上後它會給你個網址什麼的, 貼出來那個行了:)
→ wodada:http://codepad.org/04MXJH8p#comment-BkeoPFG4
→ wodada:網址如上
→ wodada:http://codepad.org/04MXJH8p
→ wodada:忘了說，我用VC++2008
→ tropical72:資料型態為double,引數為%lf非%f (%lf 都是英文字母)
→ tropical72:把所有 %f 都改成 %lf 先試試..
→ kaomark:這個switch(choice)到底是作什麼的? choice一直都是0啊
→ kaomark:然後如果你要做的事情是 選某個case跑101次
→ kaomark:那應該要在每一個case中都寫for loop
→ kaomark:如果你要的是 跑101次每次都挑一個case來做
→ kaomark:那應該要把for放在switch外面