[orf]

Shouldn't that be `a < b`?

[Ajedi32]

No, the compare function needs to be able to indicate when two values are equal, not just when one is smaller/larger than the other. Otherwise your sort function will return inconsistent results for compare(a, b) and compare(b, a) when both values are equal.

[kristianp]

So

    someList.sort((a,b) => a => b)

Would be better, because it produces a stable sort?

`a - b` produces the same results, so equally valid?

[jwilk]

(I'm assuming you meant "a >= b", not "a => b".)

That's still not a consistent comparison function, so the results are implementation-defined.

To be consistent, it's required, among others, that if cmp(a, b) == 0 then cmp(b, a) == 0. For "a >= b", this is not always true, e.g.:

  » cmp(0, 1) == 0
  true
  » cmp(1, 0) == 0
  false

[Ajedi32]

No. `(a,b) => a => b` cannot produce a stable sort in all scenarios. Would theoretically work fine for integers since there's no difference between any two instances of the same number, but consider:

  let arr = [
    {number: 2, name: 'a'},
    {number: 1, name: 'b'},
    {number: 1, name: 'c'},
  ]

Now assume we want to sort arr with a compare function, and our sort function expects the compare function to return a boolean, not an integer.

    let compare = (a,b) => a.number => b.number
    compare(arr[0], arr[1]) //=> true, so the object named 'a' comes after 'b'
    compare(arr[1], arr[2]) //=> true, so the object named 'b' comes after 'c'?

See the problem?

[52-6F-62]

That will give you a sequence of decreasing numbers. `a-b` will work just as `a > b`.

If `a-b` returns a negative number, the result will be treated as -1. If the numbers are the same, then you end up with 0, and so on.

I don't know what's faster in that case.

[jwilk]

a > b is not a consistent comparison function. The behavior of sort() with such a function is implementation-defined.

[52-6F-62]

Agreed. I wouldn't use it the way shown here. I would make sure to return a hard negative, zero, or one in any case.

I was surprised when I read into how many different ways `Array.prototype.sort()` is implemented across environments.