Question

i have two double arrays, let’s say A and B. i want to compare their results to 7 significant digits. will the following be correct to make the comparison?

k = pow(10,7);
for(...)
{
 if(((int)A[i]*k)!=((int)B[i]*k))
 {
  ...
 }
}

Answer 1

No, this will not work.

The type cast operator has higher precedence than the multiplication operator. This means that A[i] and B[i] will be cast to integers (and be truncated) before being multiplied by 1e7. 2.25 and 2.5 will end up being equal to your code. You can fix that by putting the multiplication in parentheses: (int)(A[i]*k)

Also, since you’re relying on truncation instead of rounding, you may end up with incorrect results (depending on what you’re expecting). 1.0e-7 and 1.9e-7 will be equal (1 == 1), while 1.9e-7 and 2.1e-7 will not (1 != 2). I suggest finding a function that will round properly with the behavior you desire.

Also, your comparison does not deal with significant digits, it simply changes the value of the exponent. In the above examples, there are only 2 significant digits, however your code would only compare one of those digits because the value of the exponent is -7.

Here is some code that does what you want:

//create integer value that contains 7 significant digits of input number
int adjust_num(double num) {
    double low_bound = 1e7;
    double high_bound = low_bound*10;
    double adjusted = num;
    int is_negative = (num < 0);
    if(num == 0) {
        return 0;
    }
    if(is_negative) {
        adjusted *= -1;
    }
    while(adjusted < low_bound) {
        adjusted *= 10;
    }
    while(adjusted >= high_bound) {
        adjusted /= 10;
    }
    if(is_negative) {
        adjusted *= -1;
    }
    //define int round(double) to be a function which rounds
    //correctly for your domain application.
    return round(adjusted);
}

...

if(adjust_num(A[i]) == adjust_num(B[i])) {
    ...
}