2013-11-05 68 views
0

我對C語言還是一個新東西,我並沒有真正的背景知識。我使用晶圓廠功能,然後我得到了EXC_BAD_ACCESS。誰能幫我 ?exc_bad_access,xcode,C

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

// declare all the variables needed 
int a, b, d, i, j, column, row, node, counter1, counter2, V0; 
double volt, C, h, resi, odd, even, alpha, simpson; 
double V[1000][1000], resid[1000][1000], dif_V[1000]; //1000 is chosen to get better value 

int main(void) 
{ 
    // insert code here... 
    // obtaining values of variables 
    V0 = 100; 

    printf("The potential V0 may be taken as 100\n"); 
    printf("Enter the values for dimension\n"); 

    printf("1.Width of capacitor, a="); 
    scanf("%d", &a); 

    printf("\n 2.Height of Capacitor, b="); 
    scanf("%d", &b); 

    printf("\n 3.Length of middle plate, d="); 
    scanf("%d", &d); 

    printf("\n 4.Length of square mesh grid, h="); 
    scanf("%lf", &h); 

    printf("\n 5.Value of the relaxation factor, alpha="); 
    scanf("%lf", &alpha); 

    column= a/(2*h)+1; //calculate the number of columns 
    row= b/(2*h)+1;  //calculate number of rows 
    node= d/(2*h)+1;  //calculate number of nodes 

    counter1=1; 
    counter2=1; 

    //counter1 and counter2 allows the data to be printed for each 10 scans 

    //define boundary conditions 
    //all values are set to zero 

    for(i=1;i<=column;i++) 
    {  
     for(j=1; j<=column;j++) 
      V[i][j] = 0; 
    } 

    //set middle nodes to 100 Volts 
    for(i=1;i<=node;i++) 
     V[i][j] = V0; 

    printf("\n No of scans\t"); 
    printf("Potential at the middle\t\t"); 
    printf("Maximum residual\n"); 

    do 
    { 
     //calculation of potential for every node using the 5 points formula 

     resi=0; 

     for (i=1;i<=node;i++) 
     { 
      for(j=2;j<row;j++) 
      { 
       if(i==1) 
       { 
        V[i-1][j]=V[i+1][j]; 
        //apply the 'fictitious' nodes formula for nodes at boundary 
       } 

       volt=V[i][j]; 
       //stores previous value of voltage in 'volt' 

       V[i][j]=0.25*(V[i-1][j] + V[i][j-1] + V[i][j+1] + V[i+1][j]); 
       //apply 5 points formula 

       resid[i][j]= V[i][j]-volt; 
       //calculate the residual 

       V[i][j]= volt + alpha*resid[i][j]; 
       //applying SOR scheme 
      } 
     } 

     for(i=node+1; i<column;i++)   
     { 
      for(j=1; j<row;j++)    
      { 
       if(j==1)     
       {     
        V[i-1][j]=V[i+1][j];    
       }    
       volt=V[i][j]; 
       //stores previous value of voltage in 'volt    

       V[i][j]=0.25*(V[i-1][j] + V[i][j-1] + V[i][j+1] + V[i+1][j]); 
       //apply 5 points formula 

       resid[i][j]= V[i][j]-volt; 
       //calculate the residual 

       V[i][j]= volt + alpha*resid[i][j]; 
       //applying SOR scheme 
      } 
     } 

     //compare the residuals 
     //the fabs functions compute the absolute value of a floating point number 

     for(i=1;1<=node;i++)      
     { 
      for(j=2;j<row;j++)     
      { 
       if(fabs(resid[i][j])>resi)      
       { 
        resi=fabs(resid[i][j]);      
       } 
      } 
     }  

     for(i=node+1;i<column;i++)   
     { 
      for(j=2;j<row;j++)     
      { 
       if(fabs(resid[i][j])>resi)     
       { 
        resi=fabs(resid[i][j]);      
       } 
      }    
     } 

     counter1+=1; 
     counter2+=1; 

     if(counter2==100)   
     {    
      printf("\t %3d\t\t",counter1); 
      printf("%lf\t\t\t", V[(column-1)/2][(row-1)/2]); 
      printf("%lf\n", resi); 

      counter2=0;   
     } 
    } 

    while(resi>0.0001 != counter1>(2*column*row)); 

    printf("Number of scans = %d\n", counter1); 

    //calculation of dv/dy  
    for (i=1;i<=node;i++) 
     dif_V[i]= (4*V[i][2] - 3*V[i][1] - V[i][3])/(2*h); 

    odd = 0; 
    even = 0; 

    for(i=3; i<=node; i=i+2)   
     odd= odd + dif_V[i]; 

    for(i=2;i<=node;i=i+2) 
     even= even + dif_V[i]; 

    simpson= (h/3)*(dif_V[1] + dif_V[node] + 4*even +2*odd); 

    C=(-0.04)*simpson; 

    printf("The capacitance per unit length = %lf", C); 

}