3 ms·

Shouldn't that be `a < b`?

• 52-6F-62 8y ago
  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 8y ago
    a > b is not a consistent comparison function. The behavior of sort() with such a function is implementation-defined.
    • 52-6F-62 8y ago
      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.
• Ajedi32 8y ago
  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 8y ago
    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?
    • Ajedi32 8y ago
      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?
    • jwilk 8y ago
      (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