I believe what is happening is that the literal 1 is being implicitly typecast to the Int? type by the comparison to nil. For those who aren’t used to Swift, I’ll explain a little further. Swift has a concept called “optionals”, which can either have a value or be nil. (For anyone familiar with Haskell, this is basically the Maybe monad.) It’s illegal to assign nil to a variable that wasn’t explicitly defined as optional, so let i: Int = nil will be rejected by the compiler. This allows for several benefits which are out of the scope of this answer, and it’s a rather clever way to do it.
What’s happening here, though, is that the literal 1 is a valid value of several types: Int, Int32, Int64, UInt32, UInt64, etc., etc., etc. And it’s also a valid value of the optional versions of those types: Int?, Int32?, etc.
So when the Swift compiler sees a comparison between a literal value and nil, it tries to find a type that both these values would be valid for. 1 is a valid value of the Int? type, and nil is also a valid value of the Int? type, so it applies the comparison operator with the type signature (Int?, Int?) -> Bool. (That’s the comparison operator that takes two Int? values and returns a Bool). That operator’s rules say that nil values sort lower than anything else, even Int.min, and so you get the result seen in the OP’s question.