Exhaustive condition of switch case in Swift

Question

The Apple documentation says

Every switch statement must be exhaustive. That is, every possible value of the type being considered must be matched by one of the switch cases.

So in new Xcode I have placed a code like this

println(UInt16.min); // Output : '0' println(UInt16.max); // Output : '65535'  var quantity : UInt16 = 10;  switch quantity { case 0...65535: //OR case UInt16.min...UInt16.max:     println(); default:     println(); } 

Now if i remove the default section I get a compiler error:

Switch must be exhaustive
Do you want to add missing cases? Fix

So my question is for a case that I have mentioned as case 0...65535: have I not mentioned all the case values for an UInt16 ?? But still I am getting an error ?? Why am I getting this error, Did i miss something ??

Answer 1

Swift only truly verifies that a switch block is exhaustive when working with enum types. Even a switching on Bool requires a default block in addition to true and false:

var b = true switch b { case true:  println("true") case false: println("false") } // error: switch must be exhaustive, consider adding a default clause 

With an enum, however, the compiler is happy to only look at the two cases:

enum MyBool {     case True     case False }  var b = MyBool.True switch b { case .True:  println("true") case .False: println("false") } 

If you need to include a default block for the compiler's sake but don't have anything for it to do, the break keyword comes in handy:

var b = true switch b { case true:  println("true") case false: println("false") default: break }