Compare 2 double numbers with acceptable error - always shows "not equal"

Question

Q: How can I check if two floating point numbers are equal?

A: You can do something like this:

#define EPSILON 0.00000001
int compare(double num1, double num2, double error)
{
    if(fabs(num1 - num2) < EPSILON)
        return 1;
    else
        return 0;
}

Q: How can I check if two floating point numbers are equal with some acceptable error? Meaning, I have two numbers, a = 9.2 and b = 9.7. When I set my error = 0.7 then I can consider a and b equal. (Its also true for a = 9.2 and b = 9.9 but false for a = 9.2 and b = 10.0 when the error is 0.7).

A: I tried this but every time (no matter how error look like), it always shows 0:

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

#define EPSILON 0.00000001

static int compare(double num1, double num2, double error)
{
    if((fabs(num1 - num2) < EPSILON - error) || (fabs(num1 - num2 + error) < EPSILON + error))
        return 1;
    else
        return 0;
}

static int areEqual(const double *x, int size, double error)
{
    int i;
    for (i = 0; i < size - 1; i++)
        if (!compare(x[i], x[i + 1], error))
            return 0;
    return 1;
}

int main(int argc, char **argv)
{

    double tab[] = {9.2, 9.7, 9.3, 9.6, 9.4, 10.0, 9.1, 9.7};
    double error = 0.9;
    const int N = 10;

    printf("%d\n", areEqual(tab, N, error));


    return 0;
}

Tried also do it this way:

static int compare2(double num1, double num2, double error)
{
    if(fabs(num1 - num2) <= error)
        return 1;
    else
        return 0;
}

But shows 0 too.

EDIT:

Finally, did it! Working code:

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

#define EPSILON 0.00000001

static void checkFabs()
{
    printf("%d\n", fabs(-0.7) < 1.0);
    printf("%d\n", fabs(-0.75) < 1.0);
    printf("%d\n", fabs(-0.71) < 1.0);

    printf("%d\n", fabs(0.5) < 0.9);
    printf("%d\n", fabs(0.6) < 0.9);
    printf("%d\n", fabs(0.3) < 0.9);
    printf("%d\n", fabs(0.2) < 0.9);
    printf("%d\n", fabs(0.8) < 0.9);
    printf("%d\n", fabs(0.1) < 0.9);
    printf("%d\n", fabs(0.9) <= 0.9);
}

static int compare(double num1, double num2, double error)
{
    if(fabs(num1-num2) <= error + EPSILON)
        return 1;
    else
        return 0;
}

static int areEqual(const double *x, int size, double error)
{
    int i;
    for (i = 0; i < size - 1; i++)
        if (!compare(x[i], x[i + 1], error))
            return 0;
    return 1;
}

int main(int argc, char **argv)
{

    double tab[] = {9.2, 9.7, 9.3, 9.6, 9.4, 9.9, 9.1, 9.7};
    double error = 0.9;
    const int N = 8;

    printf("%d\n", areEqual(tab, N, error));

    return 0;
}

Answer 1

There are 8 elements in your array, not 10. Here is the compare function that you need:

static int compare(double num1, double num2, double error)
{
    if(fabs(num1 - num2) < error + EPSILON)
        return 1;
    else
        return 0;
}

Answer 2

EPSILON - error in your code is a negative number!

Just compare it to error only.

Answer 3

don't define epsilon and compare using DBL_EPSILON defined in float.h header.

you have missed the error.

its actually subtracting last value with 0. so its above relative error.

checkout size.