Comparing two arrays in C, element by element

Question

I have been cracking my head at achieving something very simple in C in order to make my one of the programs (not written by me) in our computational physics project more dy

Answer 1

Your best bet is to rewrite it as a function that returns true or false (1 or 0):

int compareArrays(double a[], double b[], int n) {
  int ii;
  for(ii = 1; ii <= n; ii++) {
    if (a[ii] != b[ii]) return 0;
    // better:
    // if(fabs(a[ii]-b[ii]) < 1e-10 * (fabs(a[ii]) + fabs(b[ii]))) {
    // with the appropriate tolerance
  }
  return 1;
}

Note that it is usually bad practice to compare doubles for equality - you are better off comparing their difference, and making sure the absolute value is less than some tolerance.

Also note you are comparing elements 1 through n - C arrays start at 0 though.

You would use the above with

if (compareArrays(a, a_tmp, N)) {

where the value N is #define'd per your question.

If you want to be "clever" and avoid a loop, you can write the following - it will stop ("short-circuiting") as soon as you reach the right number of comparisons. It is still a Bad Idea to compare doubles for equality but I will leave that for another time (see comment in code above for a solution).

if(a[1]==a_temp[1] && (2 > N || (a[2]==a_temp[2] && (3 > N || (a[3]==a_temp[3]))))) {

This makes the "and the rest" true as soon as you have compared the right number of terms - so it will stop evaluating terms (as you need). I am not convinced this is either faster, or better code - but it is "dynamic"... You can obviously make this expression as long as you would like; I just wrote the first three terms so you get the idea. I DO NOT RECOMMEND IT.

As for the comparison of doubles, you might consider replacing

if(a == b)

with

if(closeEnough(a, b))

where you define the macro

#define closeEnough(a, b) (fabs((a)-(b)) < 1e-10 * (fabs(a) + fabs(b)))? 1 : 0

This will make sure that your doubles don't have to be "exactly equal" - depending on how you arrived at them, they will almost never be, and the relative tolerance of 1 part in 10^10 is usually plenty for most practical comparisons.