Why can’t Swift’s greater-than or less-than operators compare optionals when the equality operators can?

It makes perfect sense for the equality operator to support optionals, because it’s absolutely clear that for any integer valued variable i:

• nil == nil
• nil != i
• i != nil
• i == i if and only if their values are the same

On the other hand, it’s not clear how comparison to nil should act:

Is i less than nil?

• If I want to sort an array so that all the nils come out at the end, then I would want i to be less than nil.
• But if I want to sort an array so that all the nils come out at the start, then I would want i to be greater than nil.

Since either of these are equally valid, it wouldn’t make sense for the standard library to favor one over the other. It’s left to the programmer to implement whichever comparison makes sense for their use-case.

Here’s a toy implementation that generates a comparison operator to suite either case:

func nilComparator<T: Comparable>(nilIsLess: Bool) -> (T?, T?) -> Bool {
    return {
        switch ($0, $1) {
            case (nil, nil): return false
            case (nil, _?): return nilIsLess
            case (_?, nil): return !nilIsLess
            case let (a?, b?): return a < b
        }
    }
}

let input = (0...10).enumerated().map {
    $0.offset.isMultiple(of: 2) ? Optional($0.element) : nil
}

func concisePrint<T>(_ optionals: [T?]) -> String {
    return "[" + optionals.map { $0.map{ "\($0)?" } ?? "nil" }.joined(separator: ", ") + "]"
}

print("Input:", concisePrint(input))
print("nil is less:", concisePrint(input.sorted(by: nilComparator(nilIsLess: true))))
print("nil is more:", concisePrint(input.sorted(by: nilComparator(nilIsLess: false))))

Output:

Input: [0?, nil, 2?, nil, 4?, nil, 6?, nil, 8?, nil, 10?]

nil is less: [nil, nil, nil, nil, nil, 0?, 2?, 4?, 6?, 8?, 10?]

nil is more: [0?, 2?, 4?, 6?, 8?, 10?, nil, nil, nil, nil, nil]