Pearson Coefficient calculation array

Question

I'm writing a program to calculate the pearson coefficient in c, but I'm having some trouble and I'm not sure what the problem is, here is my code:

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

#define aSize 2000000


double mean(double* mean_array){
    double mean = 0;

    for (int i=0; i<aSize; i++){
        mean = mean + mean_array[i];
    }


    mean = mean/aSize;

    return mean;

}

double stan_dev_seq(double stan_array[], double stan_mean){

    double a = 0;

    for (int i=0; i<aSize; i++){
        a = a + pow((stan_array[i]-stan_mean), 2);
    }

    a = a/aSize;

    a = sqrt(a);

    return a;
}

int pearson_seq(void){

    clock_t begin, end;
    double time_spent;

    begin = clock();

    double *a;
    a = malloc(sizeof(double)*aSize);


    double *b;
    b = malloc(sizeof(double)*aSize);

    double mean_a;
    double mean_b;

    for (int i=0; i<aSize; i++){
        a[i] = sin(i);
        b[i] = sin(i+2);    

    }

    mean_a = mean(a);
    mean_b = mean(b);



    double stan_dev_a = stan_dev_seq(a, mean_a);
    double stan_dev_b = stan_dev_seq(b, mean_b);

    double pearson_numer;

    for(int i=0; i<aSize; i++){
        pearson_numer = pearson_numer + ((a[i]-mean_a)*(b[i]-mean_b));
    }

    pearson_numer = pearson_numer/aSize;

    double pearson_coef = pearson_numer/(stan_dev_a*stan_dev_b);

    printf("%s %G\n", "The Pearson Coefficient is: ", pearson_coef);

    end = clock();

    time_spent = (double)(end - begin) / CLOCKS_PER_SEC;


    printf("%f %s\n", end, "ms");
    printf("%f %s\n", begin, "ms");
    printf("%f %s\n", time_spent, "ms");



    free(a);
    free(b);

    return 0;
}

int main(void) {

    pearson_seq();

return 0;
}

If I run the program I get an incorrect value for the coefficient then a seg fault, which really makes no sense to me.

If I print the size of the arrays I get values in no way related to what the array size should be.

Any help would be appreciated.

In addition to the `sizeof` error, you should look into the format specifiers. You need `%zu` to print `size_t`, the result type of `sizeof`, and I believe the seg fault occurs when you try to print `clock_t` with `%f`. (Compiler warnings should help you to detect mismatched format specifiers and arguments.) — M Oehm, Nov 02 '15 at 12:56
the posted code does not cleanly compile!! the compiler raises two problem statements one about ` printf("%f %s\n", end, "ms");` and one about `printf("%f %s\n", begin, "ms");`. When compiling, always enable all warnings (for gcc, at a minimum use: `-Wall -Wextra -pedantic`) then fix those warnings. — user3629249, Nov 03 '15 at 23:51
Always check (!=NULL) the returned value from `malloc()` to assure the operation was successful — user3629249, Nov 03 '15 at 23:58
Note: `clock_t` is (depending on the underlying clock) a 32 or 64 bit unsigned int, not a float. — user3629249, Nov 03 '15 at 23:59

score 0 · Answer 1 · answered Nov 02 '15 at 12:53

0

for (int i=0; i<sizeof(mean_array); i++){

Arrays decay to pointers when passed to a function, so sizeof(mean_array) is equivalent to sizeof(double) which is not what you want. Pass the array size in an additional function argument.

answered Nov 02 '15 at 12:53

Magisch

7,312
9
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Hi, thank you, I've edited my code to what's shown above, but I still get the seg fault. – user2320239 Nov 02 '15 at 12:56
@user2320239 this is not the only instance of your misusing of passed array indexes. Fix the others to. – Magisch Nov 02 '15 at 12:57

score 0 · Answer 2 · edited May 23 '17 at 10:27

0

For me, crash occurs in printf statements because of using %f for clock_t type. According to the link here, there may be no specific modifier to use for clock_t. Instead you can typecast the value and print as below.

printf("%f %s\n", (float) end, "ms");
printf("%f %s\n", (float) begin, "ms");

edited May 23 '17 at 10:27

Community

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answered Nov 02 '15 at 13:00

user1969104

2,340
14
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Pearson Coefficient calculation array

2 Answers2