JavaScript sort with seemingly inconsistent results

Question

I'm trying to sort an array of numbers, but one value should be prioritized above all others, meaning it should always come out at the end of the sorted array.

I was trying playing around, trying to find a suiting sort function and stumbled upon results I don't understand.

You can see my sort function, and it's exactly the same in both cases. From my understanding, it should return 1 if a is bigger than b. Shouldn't having a fixed return value for a = 2 make 2 the "biggest" number? It seems to work with the first array, but not with the second one, so the results depend on the input array. I tried this with a lot of different inputs and I can't find a pattern for when it works and when it doesn't.

That sort comparator function is inherently invalid. You must return a consistent comparison result for any pair of elements, and that must take into account that "a" and "b" may be the same two elements you've compared before but in reverse order. — Pointy
– Pointy, Commented Apr 19, 2021 at 15:17

Drdilyor · Accepted Answer · 2021-04-19 15:28:38Z

1

You should check if b is 2 as well, this works:

console.log([5, 4, 3, 2, 1].sort((a, b) => {
  if (a === 2) return 1
  if (b === 2) return -1
  return a - b
}))

edited Apr 19, 2021 at 15:28

answered Apr 19, 2021 at 15:27

Drdilyor

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