Median calculation from float array in C

Question

I have a program which reads floating point numbers from a .txt file and puts them into an array but I have a problem with calculating median. Everything is working perfectly except from this. What am I doing wrong?

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


int compare (const void * a, const void * b)
{
  float fa = *(float*) a;
  float fb = *(float*) b;
  return (fa > fb) - (fa < fb);
}



//median calculations//
float median1(float[],int); 
float median1(float array[],int n) 
{  
qsort(array, n, sizeof(float), compare);

if(n%2==0)  
    return (array[n/2]+array[n/2-1])/2;  
else  
    return array[n/2];  
}



float x,l=~(257<<23),a,s,t,median;
main(int n,char**f)

{
char fname[20];
int i;
a=-l;

printf("Please type the file name with an extension (.txt)\n");
scanf("%s", fname);


f=fopen(fname,"r");
for(n=0;fscanf(f,"%f",&x)>0;n++,s+=x,x<l?l=x:0,x>a?a=x:0,t+=x*x);

float array[n];

fseek (f, 0, SEEK_SET);

for (i=0; i<n; i++) 
        {
            fscanf (f, "%f", &(array[n]));
        }

median=median1(array,n);  

printf("Sample size = %d\n", n);
printf("Minimum = %f\n", l);
printf("Maximum = %f\n", a);
printf("Mean = %f\n", s/n);
printf("Median = %f\n",median);
printf("Standard deviation = %f\n", sqrtf(t/n));

return 0;
}

Could you describe what's not working. Is it giving the wrong value? Try to step through it with a debugger. — MicroVirus
– MicroVirus, Commented Jan 26, 2014 at 2:17
Why would you have a collection of global variables like x, l, etc? Make them local. I'm puzzled about the initializer for l — quite honestly, I've no idea what it's supposed to do with bit operations on an integer assigned to a float. You'd also lose any marks I was giving out for the line for(n=0;fscanf(f,"%f",&x)>0;n++,s+=x,x<l?l=x:0,x>a?a=x:0,t+=x*x);. That is diabolical — ridiculous — code. — Jonathan Leffler
– Jonathan Leffler, Commented Jan 26, 2014 at 2:20
@xLokos It's high enough so long as no one tries to input something higher. — user3072164
– user3072164, Commented Jan 26, 2014 at 2:41

score 4 · Accepted Answer · 2014-01-26 02:24:34Z

4

fscanf (f, "%f", &(array[n]));

should be

fscanf (f, "%f", &(array[i]));

You are only writing to one array element and that one is out-of-bound.

Even if this wouldn't result in undefined behavior, you would still work with garbage values later on.

See @JonathanLeffler's comment for some further remarks on your code.

edited Jan 26, 2014 at 2:24

answered Jan 26, 2014 at 2:18

user3072164

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13 Comments

Jonathan Leffler Over a year ago

The first read counts how many values there are. This is used to allocate the VLA float array[n]; of the correct size.

user3072164 Over a year ago

@JonathanLeffler Ah ok, I was confused because n is for some reason the first parameter to main.

Jonathan Leffler Over a year ago

You're excused for being confused; I hadn't even noticed that. Gruesome doesn't really begin to describe some of the coding feats in here. Notice that f is the other argument to main(int n, char **f), but is also used as a FILE * in f = fopen(fname, "r"); — there's a serious case of not paying attention to compiler warnings here!

user3072164 Over a year ago

@xLokos You don't use them, so either remove them (int main(void)) or give them the usual names argc and argv. Also they are typically not written to. stackoverflow.com/questions/9356510/int-main-vs-void-main-in-c

tabstop Over a year ago

You only get about six sigfigs from a float, so -321.653992 is most likely the closest representable value to -321.654.

|

Community · Accepted Answer · 2017-05-23 12:03:46Z

0

This is a cleaned up version of your code…but select Nabla's answer.

It compiles cleanly with:

gcc -O3 -g -std=c11 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes \
    -Wold-style-definition -Werror med.c -o med

It shows that you were not dreadfully far off. I have not fixed the initializer for l. I did ensure that s and t are both zeroed before you start accumulating values in them. The code does not check that fscanf() works on the second pass. I've not fixed it to read from standard input or to take the file name from its arguments, either of which would make it a lot more usable.

Given input data:

1.2
3.5
2.9
4.6

it produces the output:

Sample size = 4
Minimum = 1.200000
Maximum = 4.600000
Mean = 3.050000
Median = 3.200000
Standard deviation = 3.288617

Semi-fixed code

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

int compare(const void *a, const void *b);

int compare(const void *a, const void *b)
{
    float fa = *(float *) a;
    float fb = *(float *) b;
    return (fa > fb) - (fa < fb);
}

// median calculations//
float median1(float[], int);

float median1(float array[], int n)
{
    qsort(array, n, sizeof(float), compare);

    if (n % 2 == 0)
        return (array[n / 2] + array[n / 2 - 1]) / 2;
    else
        return array[n / 2];
}

int main(void)
{
    float x;
    float l = ~(257 << 23);
    float a;
    float s = 0.0;
    float t = 0.0;
    float median;
    int n;
    FILE *f;
    char fname[20];
    int i;
    a = -l;

    printf("Please type the file name with an extension (.txt)\n");
    scanf("%s", fname);


    f = fopen(fname, "r");
    if (f == 0)
    {
        fprintf(stderr, "Failed to open file %s for reading\n", fname);
        return 1;
    }

    for (n = 0; fscanf(f, "%f", &x) > 0; n++)
    {
        s += x;
        if (x<l) l = x;
        if (x>a) a = x;
        t += x * x;
    }

    float array[n];

    fseek(f, 0, SEEK_SET);

    for (i = 0; i < n; i++)
    {
        fscanf(f, "%f", &(array[i]));
    }

    fclose(f);

    median = median1(array, n);

    printf("Sample size = %d\n", n);
    printf("Minimum = %f\n", l);
    printf("Maximum = %f\n", a);
    printf("Mean = %f\n", s / n);
    printf("Median = %f\n", median);
    printf("Standard deviation = %f\n", sqrtf(t / n));

    return 0;
}

edited May 23, 2017 at 12:03

CommunityBot

11 silver badge

answered Jan 26, 2014 at 2:42

Jonathan Leffler

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8 Comments

xLokos Over a year ago

I can't say it doesn't look much neater. But still even with this code and input 123.456 999.999 564.312 -654.321 7894.32 it produces output

Sample size = 5 Minimum = -654.320984 Maximum = 7894.319824 Mean = 1785.553101 Median = 564.312012 Standard deviation = 3580.003174

Jonathan Leffler Over a year ago

That's what I get too from your data:

Sample size = 5 Minimum = -654.320984 Maximum = 7894.319824 Mean = 1785.553101 Median = 564.312012 Standard deviation = 3580.003174

. It's reasonable given the inputs. Trying to get 10 significant digits from a value that can only hold 6-7 is an exercise in futility, but the default precision for %f is 6 decimal places, so that's what gets printed.

xLokos Over a year ago

So do you think this whole code will be accepted by my tutor?

xLokos Over a year ago

I just checked with my GCC and the stdev is slightly different from this in output. My Casio shows sample stdev = 3469.18976

xLokos Over a year ago

Btw thatnks for your help and time, I appreciate it.

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