Problem in output of a function

This is a discussion on Problem in output of a function within the C Programming forums, part of the General Programming Boards category; Hi, I try to make this code that normalizes an array.when I try to print the standard deviation it comes ...

  1. #1
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    Question Problem in output of a function

    Hi,
    I try to make this code that normalizes an array.when I try to print the standard deviation it comes out like this -1.#ID... and so on...
    If somebody finds the reason for it I'll appreciate the advise...
    Code:
    #include <stdio.h>
    #include <math.h>
    
    void scan_b(double b[30][30]  , int , int );
    void print_b(double b[30][30] , int  , int );
    void norm_b(double  b[30][30], int , int );
    double std_b(double b[30][30], int , int );
    double avg_b(double b[30][30], int , int );
    
    
    
    int main()
    {
    int k,l;
    double c[30][30];
    printf("please enter the number of rows and columns\n");
    scanf("%d%d" , &k, &l);
    scan_b(c ,k,l);
    printf("The original array is:");
    print_b (c,k,l);
    norm_b( c ,  k,  l);
    printf("The normalized array is:");
    print_b (c,k,l);
    printf ("the Standard deviation is: %f\n" ,std_b( c , k,  l));
    printf ("the mean value is: %f\n" ,avg_b( c ,  k,  l));
    system ("PAUSE");
    return 0;
    }
    
    
    
    void scan_b(double b[30][30] , int i, int j)
    {
    int m,n;
    printf("please enter a %d*%d array\n",i,j);
    for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   scanf("%lf" , &(b[m][n]));
    }
    
    void print_b(double b[30][30] , int i, int j)
    {
    int m,n;
    
    for (m=0 ; m<i ; m++)
           {
               printf("\n");
               for (n=0 ; n<j ; n++)
               printf("%f " , b[m][n]);
            }
            printf("\n");
    }
    
    void norm_b(double b[30][30] , int i, int j)
    {
    double sum=0,sum2=0,avg, avg2;
    int m,n;
    
    for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   sum+=(b[m][n]);
               avg=sum/(i*j);
              for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   b[m][n]=b[m][n]-avg;
                  
    for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   sum2+=pow(b[m][n],2);
                              avg2=sum2/(i*j);
                for (m=0 ; m<i ; m++)
        for (n=0 ; n<j ; n++)
            b[m][n]=b[m][n]/sqrt(avg2);
            }
    
    double avg_b(double b[30][30], int i, int j)
    {
    double sum=0,avg, avg2;
    int m,n;
    
    for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   sum+=b[m][n];
    
    avg=sum/(i*j);
    return avg;
    }
    
    double std_b(double b[30][30] , int i, int j)
    {
    double sum2,sum,std,avg,avg2,var;
    int m,n;
    for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   sum+=b[m][n];
    avg=sum/(i*j);
    for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   sum2+=pow(b[m][n],2);
    avg2=sum2/(i*j);
    for (m=0 ; m<i ; m++)
           for (n=0 ; n<j ; n++)
                   var=avg2-pow(avg,2);
    std=sqrt(var);
    return std;
    
    }

  2. #2
    and the Hat of Guessing tabstop's Avatar
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    Because you have an ... interesting ... way of computing standard deviations. You should have
    Code:
    var = avg2/(i*j)-pow(avg,2);
    It also shouldn't be inside a doubly-nested for loop.

    You also need to initialize sum to 0 before you start adding things to it.

  3. #3
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    thanks, the loop was there in mistake...
    it still doesn't solve the problem
    i already divided by i*j when i calculated avg2...

  4. #4
    Deathray Engineer MacGyver's Avatar
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    So is this like the 4th topic about this program?

  5. #5
    Registered User whiteflags's Avatar
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    I don't see where the standard deviation is actually being returned to the caller (or otherwise used after it's calculated). That is probably part of the problem.

    Also, learn how to delegate. You labored on certain functions, you should know where you can use them.

    Code:
    #include <math.h>
    
    double avg_b (double b[30][30], int i, int j)
    {
        int m, n;
        double sum = 0.0;
    
        for (m = 0; m<i; m++) {
            for (n = 0; n<j; n++) {
                sum += b[m][n];
            }
        }
        return sum / (double)(i * j);
    }
    
    double norm_b(double b[30][30], int i, int j)
    {
        double calcbuf[30][30] = {0.0,};
        double avg = 0.0;
    
        int m,n;
    
        avg = avg_b(b, i, j);
    
        for (m = 0; m<i; m++) {
            for (n = 0; n<j; n++) {
                calcbuf[m][n] = b[m][n] - avg;
                calcbuf[m][n] *= calcbuf[m][n]; /***squaring***/
            }
        }
    
        return sqrt(avg_b(calcbuf, i, j));
    }

  6. #6
    and the hat of wrongness Salem's Avatar
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    It is getting rather annoying.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

  7. #7
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    I printed the standard deviation in the main function

    Code:
    printf ("the Standard deviation is: &#37;f\n" ,std_b( c , k,  l));
    It already worked before i accidentally deleted something and I can't find what...

  8. #8
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by Livnat View Post
    thanks, the loop was there in mistake...
    it still doesn't solve the problem
    i already divided by i*j when i calculated avg2...
    I am aware of that. Since you need to divide by it twice, that means you're one short.

  9. #9
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    why do I need to divide it twice?
    var = <x^2>-<x>^2
    std=sqrt(var)
    <x^2> is avg2...

    anyways that's not the problem. I have something wrong going on with the programming it doesn't even print a number....

  10. #10
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by Livnat View Post
    why do I need to divide it twice?
    var = <x^2>-<x>^2
    std=sqrt(var)
    <x^2> is avg2...

    anyways that's not the problem. I have something wrong going on with the programming it doesn't even print a number....
    Yes, you're right there, I'm misreading my formula. So that works. You do still need to initialize sum and sum2 to 0 before you start adding things to them.

  11. #11
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    Thanks tabstop! That was the problem...
    no more topics about this one...

  12. #12
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    Quote Originally Posted by Salem View Post
    It is getting rather annoying.
    Hello Salem,
    I just saw you gave me a new way to debug in the previous topic about this non ending assignment... You're right it's smarter to have all the different problems in the same topic.
    I appreciate your help...
    -L

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