Explanation of arrayObj.sort( [sortFunction] )

Question

I know about sorting arrays, but I have the question I'm aching to find out.

For instance:

var array = [9,3,0,-2,15];

So, let's sort it:

function sortFunction(a, b){

  if(a < b) return -1; // or any number that less than zero, but why?
  if(a > b) return 1; // or any number that above zero, but why?
  if(a = b) return 0; // but why?

  // I know it may be easier a - b, I've written it for clarity

}

array.sort(sortFunction); //so we'll get correct result

How to understand why it has to return -1, 1, and 0 for sorting array?

P.S. Sorry if question seems stupid, I haven't found the answer in Google.

read this https://stackoverflow.com/questions/234683/javascript-array-sort-implementation — Vinod Louis, Oct 04 '17 at 09:59
Read the [documentation](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort#Description) and it explains it pretty clearly — George, Oct 04 '17 at 09:59
@Jarvis — Just saying you were misunderstood helps nobody. You need to clarify your problem. — Quentin, Oct 04 '17 at 12:43

score 2 · Answer 1 · answered Oct 04 '17 at 10:00

From MDN:

If compareFunction(a, b) is less than 0, sort a to an index lower than b, i.e. a comes first.

If compareFunction(a, b) returns 0, leave a and b unchanged with respect to each other, but sorted with respect to all different elements. Note: the ECMAscript standard does not guarantee this behaviour, and thus not all browsers (e.g. Mozilla versions dating back to at least 2003) respect this.

If compareFunction(a, b) is greater than 0, sort b to a lower index than a.

There are three outcomes, so three possible return values.

I've read it, I've already known this info, I think, I was misunderstood — Jarvis, Oct 04 '17 at 12:28

Explanation of arrayObj.sort( [sortFunction] )

1 Answers1