I was working with optional in switch which I noticed switch cannot accept nil as a case!
func test(value: Bool?) { switch value { case true: print("true") case false: print("false") case nil: print("nil") } } Error: Switch must be exhaustive
then I used default for solving the issue, but I am looking to learn why nil cannot work in first code?
I think this is just an incorrect diagnostic. You can add a default case, and this code will work, but I'm actually surprised that works.
You see, the value you're switching on is a Bool?, but 2 of the values you're comparing it against are true and false. The typical approach for this is to use pattern matching:
func test(value: Bool?) { switch value { case .some(true): print("true") case .some(false): print("false") case .none: print("nil") } } This works correctly, but it's a bit wordy. Luckily, Swift has a syntactic sugar for pattern matching against optionals, ? and nil:
func test(value: Bool?) { switch value { case true?: print("true") case false?: print("false") case nil: print("nil") } } Both of these snippets compile and work fine. It does open a new question as to why your original cases worked, if only you added the default.
I suggest you post about this on Swift forums, to get more input from the compiler nerds :)
An Optional is an enum with two cases, one which has the value associated: either .some(value) or .none
The correct way to use a switch for pattern matching an optional is:
switch optional { case .some(let value): // do your stuff with value case .none: // do your stuff when the optional is nil } Alternatively you could also use map on an optional to execute a closure when its value is not nil:
optional.map { value in // do your stuff with the value } With this approach the body closure passed to the map function only executes when the optional is not nil at runtime, otherwise it doesn’t executes.
Bool