Check Availability in Switch Statement

Can Swift switch statements have another switch in a case?

Of course it's possible

enum Alphabet {
  case Alpha, Beta, Gamma
}

enum Disney {
  case Goofy, Donald, Mickey
}

let foo : Alphabet = .Beta
let bar : Disney = .Mickey

switch foo {
case .Alpha, .Gamma: break
case .Beta:
  switch bar {
  case .Goofy, .Donald: break
  case .Mickey: print("Mickey")
  }
}

Swift range check in switch

It is not possible to create a range in Swift that is open on the low end.

This will accomplish what you intended:

switch value {
case 0...f1:
    print("one")
case f1...(f1 + f2):
    print("two")
case (f1 + f2)...1:
    print("three")
default:
    print("four")
}

Since switch matches the first case it finds, you don't have to worry about being non-inclusive on the low end. f1 will get matched by the "one" case, so it doesn't matter that the "two" case also includes it.

If you wanted to make the first case exclude 0, you could write it like this:

case let x where 0...f1 ~= x && x != 0:

or

case let x where x > 0 && x <= f1:

C++ Usage of switch statement

Question 1: Depends on the compiler. C++ standard does not require that a jump table be set up.

In many cases, especially with small number of sparse cases, GCC, MSVC and other compilers will do clause-by-clause check (as if it were an if statement). To give an example, suppose your cases were 1, 15, and 1000000. It would not be efficient code-wise to do a direct jump.

gcc has the option -fno-jump-tables to force it to build the equivalent if-else list.

Question 2: The break statement is not required for the last clause. It should be omitted if execution should flow down.

switch-case statement without break

The break acts like a goto command. Or, as a better example, it is like when using return in a void function. Since it is at the end, it makes no difference whether it is there or not. Although, I do like to include it.

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